NCERT Class 10 Maths Chapter 6 Exercise 6.3  Free PDF Download
Math can be a complicated subject for some students. It is especially hard for students that are preparing for their 12th board exam because of the added pressure. The chapter Application of Derivatives is seen to be avoided by most students during there because of how complex and hard to comprehend the chapter is to study. This exercise is explained in a way in which students can understand the chapter more easily. Application of Derivatives is an important chapter for students who plan on studying math in their higher studies.
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Application of Derivatives can be very confusing and time consuming for students so it is important that students be able to learn the chapter in a way that is easy for them to understand. Here in this article the notes that are provided are for free and easily accessible at all times. Here every sum of exercise 6.3 of the NCERT book is solved and explained in the easiest way possible for students to understand. Students must be aware of the basic concepts of tangents and derivatives before starting this chapter, without having the groundwork required for this chapter it will be very hard to solve the chapters sums.
Access NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives
Exercise 6.3
1. Find the slope of the tangent to the curve $y=3{{x}^{4}}4x$ at $x=4$.
Ans: The given curve is $y=3{{x}^{4}}4x$
Then, the slope of the tangent to the given curve at $x=4$ is calculated
as \[{{\frac{dx}{dy}}_{x=4}}=12{{x}^{3}}{{4}_{x=4}}=12{{\left( 4 \right)}^{3}}4=764\].
2. Find the slope of the tangents to the curve $y=\frac{x1}{x2},x\ne 2$ at $x=10$.
Ans: The given curve is $y=\frac{x1}{x2}$.
$\therefore \frac{dx}{dy}=\frac{1}{{{\left( x2 \right)}^{2}}}$
Thus, the slope of the tangent at $x=10$ is given by
${{\frac{dx}{dy}}_{x=10}}={{\frac{1}{{{\left( x2 \right)}^{2}}}}_{x=10}}=\frac{1}{64}.$
Hence, the slope of the tangent at $x=10$ is $\frac{1}{64}$.
3. Find the slope of the tangent to curve $y={{x}^{3}}x+1$ at the point whose xcoordinate is $2$.
Ans: The given curve is $y={{x}^{3}}x+1$
$\therefore \frac{dx}{dy}=3{{x}^{2}}1$
The slope of the tangent to a curve at $\left( {{x}_{0}},{{y}_{0}} \right)is{{\frac{dx}{dy}}_{\left( {{x}_{0}},{{y}_{0}} \right)}}.$
It is given that ${{x}_{0}}=2$
Hence, the slope of the tangent at the point where the xcoordinate is
$2$ is given by,
${{\frac{dx}{dy}}_{x=2}}=3{{x}^{2}}1=11$.
4. Find the slope of the tangent to the curve $y={{x}^{3}}3x+2$ at the point whose xcoordinate is $3$.
Ans: The given curve is $y={{x}^{3}}3x+2$.
$\therefore \frac{dx}{dy}=3{{x}^{2}}3$
The slope of the tangent to a curve at $\left( {{x}_{0}},{{y}_{0}} \right)is{{\frac{dx}{dy}}_{\left( {{x}_{0}},{{y}_{0}} \right)}}$
Hence, the slope of the tangent at the point where the x coordinate is $3$ is given by ${{\frac{dx}{dy}}_{x=3}}=3{{x}^{2}}3=24$.
5. Find the slope of the normal to the curve $x=a{{\cos }^{3}}\theta ,y=a{{\sin }^{3}}\theta $ at $\theta =\frac{\pi }{4}$.
Ans: It is given that $x=a{{\cos }^{3}}\theta ,y=a{{\sin }^{3}}\theta $
$ \therefore \frac{dx}{d\theta }=3a{{\cos }^{2}}\theta \left( \sin \theta \right)=3a{{\cos }^{2}}\theta \sin \theta $
$ \frac{dy}{d\theta }=3a{{\sin }^{2}}\theta \left( \cos \theta \right) $
$ \therefore \frac{dx}{d\theta }=\frac{\left( \frac{dy}{d\theta } \right)}{\left( \frac{dx}{d\theta } \right)}=\frac{3a{{\sin }^{2}}\theta \cos \theta }{3a{{\sin }^{2}}\theta \sin \theta } $
$ =\frac{\sin \theta }{\cos \theta }=\tan \theta $
Therefore, the slope of the tangent at $\theta =\frac{\pi }{4}$ is
${{\frac{dx}{dy}}_{\theta =\frac{\pi }{4}}}=\tan \theta =\tan \frac{\pi }{4}=1$
Hence, the slope of the normal is calculated as
$\frac{1}{slope\left( \tan gent \right)}=\frac{1}{1}=1$.
6. Find the slope of the normal to the curve $x=1a\sin \theta ,y=b{{\cos }^{2}}\theta $ at $\theta =\frac{\pi }{2}$.
Ans: It is given, that $x=1a\sin \theta ,y=b{{\cos }^{2}}\theta $
$ \therefore \frac{dx}{d\theta }=a\cos \theta $
$ \frac{dy}{d\theta }=\frac{\left( \frac{dy}{d\theta } \right)}{\left( \frac{dx}{d\theta } \right)}=\frac{2b\sin \theta \cos \theta }{a\cos \theta }=\frac{2b}{a}\sin \theta $
Therefore, the slope of the tangent at $\theta =\frac{\pi }{2}$ is given by
${{\frac{dy}{dx}}_{\theta =\frac{\pi }{2}}}=\frac{2b}{a}\sin \frac{\pi }{2}=\frac{2b}{a}$
Hence, the slope of the normal is given by
$\frac{1}{slope\left( \tan gent \right)}=\frac{1}{\left( \frac{2b}{a} \right)}=\frac{a}{2b}$.
7. Find the points at which the tangent to the curve $y={{x}^{3}}3{{x}^{3}}9x+7$ is parallel to the xaxis.
Ans: The provided curve has the equation $y={{x}^{3}}3{{x}^{3}}9x+7$
$\therefore \frac{dy}{dx}=3{{x}^{2}}6x9$
If the slope of the tangent is zero, the tangent is now parallel to the xaxis.
$ \therefore 3{{x}^{2}}6x9=0 $
$ \Rightarrow {{x}^{2}}2x3=0 $
$ \Rightarrow \left( x3 \right)\left( x+1 \right)=0 $
$ \Rightarrow x=3,1 $
When, $x=3,y={{\left( 3 \right)}^{2}}9\left( 3 \right)+7=20$
When $x=1,y={{\left( 1 \right)}^{3}}3{{\left( 1 \right)}^{2}}9\left( 1 \right)+7=12.$
Hence, the points at which the tangent is parallel to the xaxis are
$\left( 3,20 \right)$ and $\left( 1,12 \right)$.
8. Find a point on the curve $y={{\left( x2 \right)}^{2}}$ at which the tangent is parallel to the chord joining the points $\left( 2,0 \right)and\left( 4,4 \right).$
Ans: We know that if a tangent is parallel to the chord connecting the points $\left( 2,0 \right)and\left( 4,4 \right)$, then the slope of the tangent equals the slope of the chord.
The slope of the chord is $\frac{40}{42}=2$.
Now, the slope of the tangent to the given curve at a point$\left( x,y \right)$ is
given by $\frac{dy}{dx}=2\left( x2 \right)$.
Since the slope of the tangent = slope of the chord, we have:
$ 2\left( x2 \right)=2 $
$ \Rightarrow x2=1\Rightarrow x=3 $
When $x=3,y={{\left( 32 \right)}^{2}}=1$
Hence, the required point is $\left( 3,1 \right)$.
9. Find the point on the curve $y={{x}^{3}}11x+5$ at which the tangent is $y=x11$.
Ans: The equation of the given curve is $y={{x}^{3}}11x+5$.
The equation of the tangent to the given curve is given as $y=x11$
which is of the form ($y=mx+c$)
Slope of the tangent $=1$
Now, the slope of the tangent to the given curve at the point $\left( x,y\right)$ is
given by
$\frac{dy}{dx}=3{{x}^{2}}11$
Then, we have:
$ 3{{x}^{2}}11=1 $
$ \Rightarrow 3{{x}^{2}}=12 $
$ \Rightarrow {{x}^{2}}=4 $
$ \Rightarrow x=\pm 2 $
When $x=2,y={{\left( 2 \right)}^{3}}11\left( 2 \right)+5=9$
When $x=2,y={{\left( 2 \right)}^{3}}11\left( 2 \right)+5=19.$
Hence, the required points are $\left( 2,9 \right)$ and $\left( 2,19\right).$
10. Find the equation of all lines having slope $1$ that are tangents to the curve $y=\frac{1}{x1},x\ne 1$.
Ans: The equation of the given curve is $y=\frac{1}{x1},x\ne 1$.
The slope of the tangents to the given curve at any point $\left( x,y \right)$ is
given by $\frac{dy}{dx}=\frac{1}{{{\left( x1 \right)}^{2}}}$
If the slope of the tangents is $1$ then we have,
$ \frac{1}{{{\left( x1 \right)}^{2}}}=1 $
$ \Rightarrow {{\left( x1 \right)}^{2}}=1 $
$ \Rightarrow x1=\pm 1 $
$ \Rightarrow x=2,0 $
When $x=0,y=1$ and when $x=2,y=1$.
Thus, the given curve having slope $1$ has two tangents.
These are passing through the points $\left( 0,1 \right)and\left( 2,1 \right)$.
Therefore, the equation of the tangent through $\left( 0,1 \right)$ is given by
$ y\left( 1 \right)=1\left( x0 \right) $
$ \Rightarrow y+1=x $
$ \Rightarrow y+x+1=0 $
Therefore, the equation of the tangent through $\left( 2,1 \right)$ is given by
$ y1=1\left( x2 \right) $
$ \Rightarrow y1=x+2 $
$ \Rightarrow y+x3=0 $
Hence, the equations of the required lines are $y+x+1=0$ and
$y+x3=0$.
11. Find the equation of all lines having slope $2$ which are tangents to the curve $y=\frac{1}{x3},x\ne 3.$
Ans: The equation of the given curve is $y=\frac{1}{x3},x\ne 3.$
The slope of the tangent to the given curve at any point $\left( x,y \right)$ is given
by, $\frac{dy}{dx}=\frac{1}{{{\left( x3 \right)}^{2}}}$
If the slope of the tangent is $2$, then we have:
$ \frac{1}{{{\left( x3 \right)}^{2}}}=2 $
$ \Rightarrow 2{{\left( x3 \right)}^{2}}=1 $
$ \Rightarrow {{\left( x3 \right)}^{2}}=\frac{1}{2} $
There is no tangent to the given curve with slope 2 because the L.H.S. is positive and the R.H.S. is negative.
12. Find the equations of all lines having slope $0$ which are tangent to the curve $y=\frac{1}{{{x}^{2}}2x+3}.$
Ans: The equation of the given curve is $y=\frac{1}{{{x}^{2}}2x+3}.$
The slope of the tangent to the given curve at any point $\left( x,y \right)$ is given by, $\frac{dy}{dx}=\frac{2\left( 2x2 \right)}{{{\left( {{x}^{2}}2x+3 \right)}^{2}}}=\frac{2\left( x1 \right)}{{{\left( {{x}^{2}}2x+3 \right)}^{2}}}$
If the slope of the tangent is $0$, then we have:
$ \frac{2\left( x1 \right)}{{{\left( {{x}^{2}}2x+3 \right)}^{2}}}=0 $
$ \Rightarrow 2\left( x1 \right)=0 $
$ \Rightarrow x=1 $
When $x=1,y=\frac{1}{2}.$
Therefore, the equation of the tangent through $\left( 1,\frac{1}{2} \right)$ is given by
$ y\frac{1}{2}=0\left( x1 \right) $
$ \Rightarrow y\frac{1}{2}=0 $
$ \Rightarrow y=\frac{1}{2} $
Hence, the equation of the required line is $y=\frac{1}{2}$.
13. Find points on the curve \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\] at which the tangents are
(i). Parallel to xaxis
(ii). Parallel to yaxis
Ans: The equation of the given curve is \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\].
On differentiating both sides for $x$ , we have:
$ \frac{2x}{9}+\frac{2y}{16}.\frac{dy}{dx}=0 $
$ \Rightarrow \frac{dy}{dx}=\frac{16x}{9y} $
(i). The tangent is parallel to the xaxis if the slope of the tangent is 0 i.e., $\frac{16x}{9y}=0$, which is possible if $x=0$.
Then, $\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1$ for $x=0$.
$ \Rightarrow {{y}^{2}}=16 $
$ \Rightarrow y=\pm 4 $
Hence, the points at which the tangents are parallel to the xaxis are
$\left( 0,4 \right)$ and $\left( 0,4 \right).$
(ii). The tangent is parallel to the y axis if the slope of the normal is $0$ which gives
(i). $ \frac{1}{\left( \frac{16x}{9y} \right)}=\frac{9y}{16x}=0 $
(ii). $ \Rightarrow y=0 $
(iii). Then, $\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1$ for $y=0.$ $\Rightarrow x=\pm 3$.
Hence, the points at which the tangents are parallel to the yaxis are $\left( 3,0 \right)$ and $\left( 3,0 \right)$.
14. Find the equations of the tangents and normal to the given curves at
the indicated points:
(i). $y={{x}^{4}}6{{x}^{3}}+13{{x}^{2}}10x+5at\left( 0,5 \right)$
(ii). $y={{x}^{4}}6{{x}^{3}}+13{{x}^{2}}10x+5at\left( 1,3 \right)$
(iii). $y={{x}^{3}}at\left( 1,1 \right)$
(iv). $y={{x}^{2}}at\left( 0,0 \right)$
(v). $x=\cos t,y=\sin t$ at $t=\frac{\pi }{4}$
Ans:
(i). The equation of the curve is $y={{x}^{4}}6{{x}^{2}}+13{{x}^{2}}10x+5$.
On differentiating with respect to x, we get:
$ \frac{dy}{dx}=4{{x}^{3}}18{{x}^{2}}+26x10 $
$ {{\frac{dy}{dx}}_{\left( 0,5 \right)}}=10 $
Thus, the slope of the tangent at $\left( 0,5 \right)$ is $10.$ The equation of the tangent id given as:
$ y5=10\left( x0 \right) $
$ \Rightarrow y5=10\left( x0 \right) $
$ \Rightarrow y5=10x $
$ \Rightarrow 10x+y=5 $
The slope of the normal at $\frac{1}{slope\left( \tan gent \right)}=\frac{1}{10}$
Therefore, the equation of the normal at $\left( 0,5 \right)$ is given as:
$ y5=\frac{1}{10}\left( x0 \right) $
$ \Rightarrow 10y50=x $
$ \Rightarrow x10y+50=0 $
(ii). The equation of the curve is $y={{x}^{4}}6{{x}^{3}}+13{{x}^{2}}10x+5.$
On differentiating with respect to $x$, we get:
$ \frac{dy}{dx}=4{{x}^{3}}18{{x}^{2}}+26x10 $
$ {{\frac{dy}{dx}}_{\left( 1,3 \right)}}=418+2610=2 $
Thus, the slope of the tangent at $\left( 1,3 \right)$ is $2$. The equation of the tangent is given as:
$ y3=2\left( x1 \right) $
$ \Rightarrow y3=2x2 $
$ \Rightarrow y=2x+1 $
The slope of the normal at $\left( 1,3 \right)$ is $\frac{1}{slope\left( \tan gent \right)}=\frac{1}{2}$
Therefore, the equation of the normal at $\left( 1,3 \right)$ is given as:
$ y3=\frac{1}{2}\left( x1 \right) $
$ \Rightarrow 2y6=x+1 $
$ \Rightarrow x+2y7=0 $
(iii). The equation of the curve is $y={{x}^{3}}$. On differentiating with respect to $x$, we get:
$ \frac{dy}{dx}=3{{x}^{2}} $
$ {{\frac{dy}{dx}}_{\left( 1,1 \right)}}=3{{\left( 1 \right)}^{2}}=3 $
Thus, the slope of the tangent at $\left( 1,1 \right)$ is $3$and the equation of the tangent is given as:
$ y1=3\left( x1 \right) $
$ \Rightarrow y=3x2 $
The slope of the normal at \[\left( 1,1 \right)\] is $\frac{1}{slope\left( \tan gent \right)}=\frac{1}{3}.$
Therefore, the equation of the normal at $\left( 1,1 \right)$ is given as:
$ y1=\frac{1}{3}\left( x1 \right) $
$ \Rightarrow 3y3=x+1 $
$ \Rightarrow x+3y4=0 $
(iv). The equation of the curve is $y={{x}^{2}}$.
On differentiating with respect to $x$, we get:
$ \frac{dy}{dx}=2x $
$ {{\frac{dy}{dx}}_{\left( 0,0 \right)}}=0 $
Thus, the slope of the tangent at $\left( 0,0 \right)$ is $0$ and the equation of the tangent is given as:
$ y0=0\left( x0 \right) $
$ \Rightarrow y=0 $
The slope of the normal at $\left( 0,0 \right)$ is $\frac{1}{slope\left( \tan gent \right)}=\frac{1}{0}$,which is not defined.
Therefore, the equation of the normal at $\left( {{x}_{0}},{{y}_{0}} \right)=\left( 0,0 \right)$ is given by $x={{x}_{0}}=0.$
(v). The equation of the curve is $x=\cos t,y=\sin t$
On differentiating with respect to $x$, we get:
$ \therefore \frac{dx}{dt}=\sin t,\frac{dy}{dt}=\cos t $
$ \therefore \frac{dy}{dx}=\frac{\left( \frac{dy}{dt} \right)}{\left( \frac{dx}{dt} \right)}=\frac{\cos t}{\sin t}=\cot t $
$ {{\frac{dy}{dx}}_{t=\frac{\pi }{4}}}=\cot t=1 $
The slope of the tangent at $t=\frac{\pi }{4}$ is $1.$
When, $t=\frac{\pi }{4},x=\frac{1}{\sqrt{2}},y=\frac{1}{\sqrt{2}}.$
Thus, the equation of the tangent to the given curve at $t=\frac{\pi }{4}i.e.,\left[ \left( \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right) \right]$ is
$ y\frac{1}{\sqrt{2}}=1\left( x\frac{1}{\sqrt{2}} \right). $
$ \Rightarrow x+y\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}=0 $
$ \Rightarrow x+y\sqrt{2}=0 $
The slope of the normal at $t=\frac{\pi }{4}$ is $\frac{1}{slope\left( \tan gent \right)}=1.$
Therefore, the equation of the normal to the given curve at $t=\frac{\pi }{4}i.e.,$at $\left[ \left( \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right) \right]$ is $y\frac{1}{\sqrt{2}}=1\left( x\frac{1}{\sqrt{2}} \right).$
$\Rightarrow x=y$
15. Find the equation of the tangent line to the curve $y={{x}^{2}}2x+7$
which is
Parallel to the line $2xy+9=0$
Perpendicular to the line $5y15x=13.$
Ans:
The equation of the given curve is $y={{x}^{2}}2x+7$
On differentiating with respect to $x$, we get:
$\frac{dy}{dx}=2x2.$
The equation of the line is $2xy+9=0$.
$2xy+9=0,y=2x+9$
This is of the form $mx+c.$
Slope of the line is $2$.
The slope of the tangent is equal to the slope of the line if a tangent is parallel to the line $2xy+9=0$.
Therefore, we have:
$ 2=2x2 $
$ \Rightarrow 2x=4 $
$ \Rightarrow x=2 $
We get, $y=44+7=7$
Thus, the equation of the equation of the tangent passing through $\left( 2,7 \right)$ is given by,
$ y7=2\left( x2 \right) $
$ y2x3=0 $
Hence, the equation of the tangent line to the given curve (which is parallel to the line $2xy+9=0$) is $y2x3=0.$
The equation of the line is $5y15x=13.$
$y=3x+\frac{13}{5}$
This is of the form $mx+c.$
Slope of the line is $3$.
The slope of the tangent is $\frac{1}{slope\left( line \right)}=\frac{1}{3}$ if a tangent is perpendicular to the line $5y15x=13,$
$ \Rightarrow 2x2=\frac{1}{3} $
$ \Rightarrow 2x=\frac{5}{3} $
$ \Rightarrow x=\frac{5}{6} $
$ Then,y=\frac{25}{36}+\frac{10}{6}+7=\frac{217}{36} $
Thus, the equation of the tangent passing through $\left( \frac{5}{6},\frac{217}{36} \right)$ is given by,
$ y\frac{217}{36}=\frac{1}{3}\left( x\frac{5}{6} \right) $
$ \Rightarrow 36y217=2\left( 6x5 \right) $
$ \Rightarrow 36y+12x227=0 $
As a result, the tangent line to the provided curve (which is perpendicular to the line $5y15x=13)$) has the equation $36y+12x227=0$.
16. Show that the tangents to the curve $y=7{{x}^{3}}+11$ at the points where $x=2$ and $x=2$ are parallel.
Ans: The equation of the given curve is $y=7{{x}^{3}}+11$.
$\therefore \frac{dy}{dx}=21{{x}^{2}}$
The slope of the tangent to a curve at $\left( {{x}_{0}},{{y}_{0}} \right)is{{\frac{dy}{dx}}_{\left( {{x}_{0}},{{y}_{0}} \right)}}$
Therefore, the slope of the tangent at the point where $x=2$ is calculated as ${{\frac{dy}{dx}}_{x=2}}=21{{\left( 2 \right)}^{2}}=84$.
It is observed that the slope of the tangents at the points where $x=2$ and $x=2$ are equal.
Hence, the two tangents are parallel.
17. Find the points on the curve $y={{x}^{3}}$ at which the slope of the tangent is equal to the y coordinate of the point.
Ans: The equation of the given curve is $y={{x}^{3}}$.
$\therefore \frac{dy}{dx}=3{{x}^{2}}$
The slope of the tangent at the point $\left( x,y \right)$ is given by,
${{\frac{dy}{dx}}_{\left( x,y \right)}}=3{{x}^{2}}$
When the slope of the tangent is equal to the ycoordinate of the point, then we have $y={{x}^{3}}$
$ 3{{x}^{2}}={{x}^{3}} $
$ {{x}^{2}}\left( x3 \right)=0 $
$ x=0,x=3 $
When $x=0$, then $y=0$ , and when $x=3$, then $y=3{{\left( 3 \right)}^{2}}=27$.
Hence, the required points are $\left( 0,0 \right)$ and $\left( 3,27 \right)$.
18. For the curve, $y=4{{x}^{3}}2{{x}^{5}},$find all the points at which the tangents pass through the origin.
Ans: The equation of the given curve is $y=4{{x}^{3}}2{{x}^{5}}$.
$\therefore \frac{dy}{dx}=12{{x}^{2}}10{{x}^{4}}$
Therefore, the slope of the tangent at a point $\left( x,y \right)is12{{x}^{2}}10{{x}^{4}}.$
The equation of the tangent at $\left( x,y \right)$ is given by,
$Yy=\left( 12{{x}^{2}}10{{x}^{4}} \right)\left( Xx \right)$ …... (1)
When the tangent passes through the origin $\left( 0,0 \right)$, then $X=Y=0.$
Therefore, equation (1) reduces to:
$ y=\left( 12{{x}^{2}}10{{x}^{4}} \right)\left( x \right) $
$ y=12{{x}^{3}}10{{x}^{5}} $
Also, we have $y=4{{x}^{3}}2{{x}^{5}}.$
$ \therefore 12{{x}^{3}}10{{x}^{5}}=4{{x}^{3}}2{{x}^{5}} $
$ \Rightarrow 8{{x}^{5}}8{{x}^{3}}=0 $
$ \Rightarrow {{x}^{5}}{{x}^{3}}=0 $
$ \Rightarrow {{x}^{3}}\left( {{x}^{2}}1 \right)=0 $
$ \Rightarrow x=0,\pm 1 $
When $x=0,y=4{{\left( 0 \right)}^{3}}2{{\left( 0 \right)}^{5}}=0$
When $x=1,y=4{{\left( 1 \right)}^{3}}2{{\left( 1 \right)}^{5}}=2$
When $x=1,y=4{{\left( 1 \right)}^{3}}2{{\left( 1 \right)}^{5}}=2$
Hence, the required points are $\left( 0,0 \right),\left( 1,2 \right)\And \left( 1,2 \right).$
19. Find the points on the curve ${{x}^{2}}+{{y}^{2}}2x3=0$ at which the tangents are parallel to the xaxis.
Ans: The given curve has the following equation: ${{x}^{2}}+{{y}^{2}}2x3=0$ On differentiating with respect to $x,$we have:
$ 2x+2y\frac{dy}{dx}2=0 $
$ \Rightarrow y\frac{dy}{dx}=1x $
$ \Rightarrow \frac{dy}{dx}=\frac{1x}{y} $
Now, if the slope of the tangent is 0, the tangents are parallel to the xaxis.
$ \therefore \frac{1x}{y}=0\Rightarrow 1x=0 $
$ \Rightarrow x=1 $
$ But,{{x}^{2}}+{{y}^{2}}2x3=0 $
$ \Rightarrow {{y}^{2}}=4,y=\pm 2 $
Hence, $\left( 1,2 \right)$ and $\left( 1,2 \right)$ are the points at which the tangents are parallel to the x axis.
20. Find the equation of the normal at the point $\left( a{{m}^{2}},a{{m}^{3}} \right)$ for the curve $a{{y}^{2}}={{x}^{3}}.$
Ans: The given curve has the following equation: $a{{y}^{2}}={{x}^{3}}$
On differentiating with respect to $x$ we get:
$ 2ay\frac{dy}{dx}=3{{x}^{2}} $
$ \Rightarrow \frac{dy}{dx}=\frac{3{{x}^{2}}}{2ay} $
The slope of a tangent to the curve at $\left( {{x}_{0}},{{y}_{0}} \right)is{{\frac{dy}{dx}}_{\left( {{x}_{0}},{{y}_{0}} \right)}}.$
This suggests that the slope of the tangent to the given curve at
$\left( a{{m}^{2}},a{{m}^{3}} \right)$ is
${{\frac{dy}{dx}}_{\left( a{{m}^{2}},a{{m}^{3}} \right)}}=\frac{3{{\left( a{{m}^{2}} \right)}^{2}}}{2a\left( a{{m}^{3}} \right)}=\frac{3m}{2}.$
Slope of normal at $\left( a{{m}^{2}},a{{m}^{3}} \right)=\frac{1}{slope\left( \tan gent \right)}=\frac{2}{3m}$
Hence, the normal at $\left( a{{m}^{2}},a{{m}^{3}} \right)$has the following equation,
$ ya{{m}^{3}}=\frac{2}{3m}\left( xa{{m}^{2}} \right) $
$ \Rightarrow 3my3a{{m}^{4}}=2x+2a{{m}^{2}} $
$ \Rightarrow 2x+3mya{{m}^{2}}\left( 2+3{{m}^{2}} \right)=0 $
21. Find the equation of the normal to the curve $y={{x}^{3}}+2x+6$ which are parallel to the line $x+14y+4=0.$
Ans: The given curve has the following equation: $y={{x}^{3}}+2x+6.$
The slope of the tangent to the given curve at any point $\left( x,y \right)$ is given by, $\frac{dy}{dx}=3{{x}^{2}}+2$
At any point, slope of the normal to the given curve
$\left( x,y \right)=\frac{1}{Slope\left( \tan gent \right)}=\frac{1}{3{{x}^{2}}+2}$.
The given line has the equation: $x+14y+4=0.$
$x+14y+4=0,y=\frac{1}{14}x\frac{4}{14}$ (which is of the form $mx+c$)
Slope of the line $=\frac{1}{14}$
If the normal is parallel to the line, then the slope of the
normal must be equal to the slope of the line.
$ \therefore \frac{1}{3{{x}^{2}}+2}=\frac{1}{14} $
$ \Rightarrow 3{{x}^{2}}+2=14 $
$ \Rightarrow {{x}^{2}}=4 $
$ \Rightarrow x=\pm 2 $
When $x=2,y=8+4+6=18$
When $x=2,y=84+6=6.$
Therefore, there are two normal to the given curve with slope $\frac{1}{14}$ and passing through the points $\left( 2,18 \right)and\left( 2,16 \right).$
Thus, the normal through $\left( 2,18 \right)$has the following equation:
$ y18=\frac{1}{14}\left( x2 \right) $
$ \Rightarrow 14y252=x+2 $
$ \Rightarrow x+14y252=0 $
And, the normal through $\left( 2,6 \right)$ has the following equation:
$ y\left( 6 \right)=\frac{1}{14}\left[ x\left( 2 \right) \right] $
$ \Rightarrow y+6=\frac{1}{14}\left( x+2 \right) $
$ \Rightarrow 14y+84=x2\backslash $
$ \Rightarrow x+14y+86=0 $
Hence, the equations of the normal to the given curve (which are parallel to the given line) are $x+14y254=0$ and $x+14y+86=0.$
22. Find the equations of the tangent and normal to the parabola ${{y}^{2}}=4ax$ at the point $\left( a{{t}^{2}},2at \right)$.
Ans: The given parabola has the following equation: ${{y}^{2}}=4ax$. On differentiating ${{y}^{2}}=4ax$ with respect to $x$, we have:
$ 2y\frac{dy}{dx}=4a $
$ \Rightarrow \frac{dy}{dx}=\frac{2a}{y} $
The slope of the tangent at $\left( a{{t}^{2}},2at \right)is{{\frac{dy}{dx}}_{\left( a{{t}^{2}},2at \right)}}=\frac{2a}{2at}=\frac{1}{t}.$
Then, the equation of the tangent at $\left( a{{t}^{2}},2at \right)$ is given by,
$\frac{1}{slope\left( \tan gent \right)}=t$
Thus, the normal at $\left( a{{t}^{2}},2at \right)$ has the following equation:
$ y2at=t\left( xa{{t}^{2}} \right) $
$ \Rightarrow y2at=tx+a{{t}^{3}} $
$ \Rightarrow y=tx+2at+a{{t}^{3}} $
23. Prove that the curves $x={{y}^{2}}$ and $xy=k$ cut at angles if $8{{k}^{2}}=1$. (Hint: Two curves intersect at right angle if the tangents to the curves at the point of intersection are perpendicular to each other.)
Ans: The equation of the given curves are given as $x={{y}^{2}}$ and $xy=k$. Putting $x={{y}^{2}}$ in $xy=k$ we get:
$ {{y}^{3}}=k\Rightarrow y={{k}^{\frac{1}{3}}} $
$ \therefore x={{k}^{\frac{2}{3}}} $
Thus, the point of intersection of the given curves is $\left( {{k}^{\frac{2}{3}}},{{k}^{\frac{1}{3}}} \right)$
Differentiating $x={{y}^{2}}$ with respect to $x$ we have:
$1=2y\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{2y}$
Therefore, the slope of the tangent to the curve
$x={{y}^{2}}at\left( {{k}^{\frac{2}{3}}},{{k}^{\frac{1}{3}}} \right)is{{\frac{dy}{dx}}_{\left( {{k}^{\frac{2}{3}}},{{k}^{\frac{1}{3}}} \right)}}=\frac{1}{2{{k}^{\frac{1}{3}}}}$
On differentiating $xy=k$ with respect to $x$, we have:
$x\frac{dy}{dx}+y=0\Rightarrow \frac{dy}{dx}=\frac{y}{x}$
Slope of the tangent to the curve $xy=k$ at $\left(\mathrm{k}^{\frac{2}{3}}, \mathrm{k}^{\frac{1}{3}}\right)$ is given by:
${{\frac{dy}{dx}}_{\left( {{k}^{\frac{2}{3}}},{{k}^{\frac{1}{3}}} \right)}}={{\frac{y}{x}}_{\left( {{k}^{\frac{2}{3}}},{{k}^{\frac{1}{3}}} \right)}}=\frac{{{k}^{\frac{1}{3}}}}{{{k}^{\frac{2}{3}}}}=\frac{1}{{{k}^{\frac{1}{3}}}}$
We know that if the tangents to the curves at the point of intersection i.e., at $\left(\mathrm{k}^{\frac{2}{3}},\mathrm{k}^{\frac{1}{3}}\right)$ are perpendicular to each other, then two curves intersect at 90 degrees angle.
This implies that the product of the tangents should be $1$ .
Thus, the given two curves will cut at right angles if the product of the
slopes of their respective tangents at $\left( {{k}^{\frac{2}{3}}},{{k}^{\frac{1}{3}}} \right)$ is $1.$
$ i.e.,\left( \frac{1}{2{{k}^{\frac{1}{3}}}} \right)\left( \frac{1}{{{k}^{\frac{1}{3}}}} \right)=1 $
$ \Rightarrow 2{{k}^{\frac{2}{3}}}=1 $
$ \Rightarrow {{\left( 2{{k}^{\frac{2}{3}}} \right)}^{3}}={{\left( 1 \right)}^{3}} $
$ \Rightarrow 8{{k}^{2}}=1 $
Hence, the given two curves cut at right angles if $8{{k}^{2}}=1.$
24. Find the equations of the tangent and normal to the hyperbola $\frac{{{x}^{2}}}{{{a}^{2}}}\frac{{{y}^{2}}}{{{b}^{2}}}=1$ at the point $\left( {{x}_{0}},{{y}_{0}} \right)$.
Ans: Differentiating $\frac{{{x}^{2}}}{{{a}^{2}}}\frac{{{y}^{2}}}{{{b}^{2}}}=1$ with respect to $x$ we have:
$ \frac{2x}{{{a}^{2}}}\frac{2y}{{{b}^{2}}}\frac{dy}{dx}=0 $
$ \Rightarrow \frac{2y}{{{b}^{2}}}\frac{dy}{dx}=\frac{2x}{{{a}^{2}}} $
$ \Rightarrow \frac{dy}{dx}=\frac{{{b}^{2}}x}{{{a}^{2}}y} $
Therefore, the slope of the tangent at $\left( {{x}_{0}},{{y}_{0}} \right)$ is ${{\frac{dy}{dx}}_{\left( {{x}_{0}},{{y}_{0}} \right)}}=\frac{{{b}^{2}}{{x}_{0}}}{{{a}^{2}}{{y}_{0}}}$
Then, the tangent at $\left( {{x}_{0}},{{y}_{0}} \right)$ has the following equation:
$ y{{y}_{0}}=\frac{{{b}^{2}}{{x}_{0}}}{{{a}^{2}}{{y}_{0}}}\left( x{{x}_{0}} \right) $
$ \Rightarrow {{a}^{2}}y{{y}_{0}}{{a}^{2}}y_{0}^{2}={{b}^{2}}x{{x}_{0}}{{b}^{2}}x_{0}^{2} $
$ \Rightarrow {{b}^{2}}x{{x}_{0}}{{a}^{2}}y{{y}_{0}}{{b}^{2}}x_{0}^{2}+{{a}^{2}}y_{0}^{2}=0 $
$\Rightarrow \frac{x{{x}_{0}}}{{{a}^{2}}}\frac{y{{y}_{0}}}{{{b}^{2}}}\left( \frac{x_{0}^{2}}{{{a}^{2}}}\frac{y_{0}^{2}}{{{b}^{2}}} \right)=0$ (On dividing both sides by ${{a}^{2}}{{b}^{2}}$ )
$\Rightarrow \frac{x{{x}_{0}}}{{{a}^{2}}}\frac{y{{y}_{0}}}{{{b}^{2}}}1=0$ (\[\left( {{x}_{0}},{{y}_{0}} \right)\]lies on the hyperbola $\frac{{{x}^{2}}}{{{a}^{2}}}\frac{{{y}^{2}}}{{{b}^{2}}}=1$ )
$\Rightarrow \frac{x{{x}_{0}}}{{{a}^{2}}}\frac{y{{y}_{0}}}{{{b}^{2}}}=1$
Now, the slope of the normal at$\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$is given by,
$\frac{1}{slope\left( \tan gent \right)}=\frac{{{a}^{2}}{{y}_{0}}}{{{b}^{2}}{{x}_{0}}}$
Hence, the normal at $\left( {{x}_{0}},{{y}_{0}} \right)$ has the following equation:
$ y{{y}_{0}}=\frac{{{a}^{2}}{{y}_{0}}}{{{b}^{2}}{{x}_{0}}}\left( x{{x}_{0}} \right) $
$ \Rightarrow \frac{y{{y}_{0}}}{{{a}^{2}}{{y}_{0}}}=\frac{\left( x{{x}_{0}} \right)}{{{b}^{2}}{{x}_{0}}} $
$ \Rightarrow \frac{y{{y}_{0}}}{{{a}^{2}}{{y}_{0}}}+\frac{\left( x{{x}_{0}} \right)}{{{b}^{2}}{{x}_{0}}}=0 $
25. Find the equation of the tangent to the curve $y=\sqrt{3x2}$ which is parallel to the line $4x2y+5=0.$
Ans: The given curve has the following equation: $y=\sqrt{3x2}$.
The tangent to the given curve at any point $\left( x,y \right)$ has the following slope: $\frac{dy}{dx}=\frac{3}{2\sqrt{3x2}}$ The given line has the following equation: $4x2y+5=0.$ $4x2y+5=0,y=2x+\frac{5}{2}$ (which is of the form $mx+c$) Slope of the line is
$ \frac{3}{2\sqrt{3x2}}=2 $
$ \Rightarrow \sqrt{3x2}=\frac{3}{4} $
$ \Rightarrow 3x2=\frac{9}{16} $
$ \Rightarrow 3x=\frac{9}{16}+2=\frac{41}{16} $
$ \Rightarrow x=\frac{41}{48} $
Now, if the slope of the tangent is equal to the slope of the line,
then the tangent to the given curve is parallel to the line $4x2y+5=0.$
$\frac{3}{2 \sqrt{3 x2}}=2$
$\Rightarrow \sqrt{3 x2}=\frac{3}{4}$
$\Rightarrow 3 x2=\frac{9}{16}$
$\Rightarrow 3 x=\frac{9}{16}+2=\frac{41}{16}$
$\Rightarrow x=\frac{41}{48}$
When $x=\frac{41}{48},y=\sqrt{3\left( \frac{41}{48} \right)2}=\sqrt{\frac{41}{16}2}=\sqrt{\frac{9}{16}}=\frac{3}{4}.$
the tangent passing through the point $\left( \frac{41}{48},\frac{3}{4} \right)$ has the following equation:
$ y\frac{3}{4}=2\left( x\frac{41}{48} \right) $
$ \Rightarrow \frac{4y3}{4}=2\left( \frac{48x41}{48} \right) $
$ \Rightarrow 4y3=\left( \frac{4841}{6} \right) $
$ \Rightarrow 48x24y=23 $
Hence, the required tangent has the following equation: $48x24y=23.$
26. The slope of the normal to the curve $y=2{{x}^{2}}+3\sin x$ at $x=0$ is
$\left( A \right)3,\left( B \right)\frac{1}{3},\left( C \right)3,\left( D \right)\frac{1}{3}$
Ans: The given curve has the following equation:
$y=2{{x}^{2}}+3\sin x$. the tangent to the given curve at $x=0$ has the following slope:
\[{{\frac{dy}{dx}}_{x=0}}={{[4x+3\cos x]}_{x=0}}=0+3\cos 0=3\]
Hence, the normal to the given curve at $x=0$ has the slope:
$\frac{1}{slope\left( \tan gent \right)}=\frac{1}{3}.$ The correct option is D.
27. The line $y=x+1$ is a tangent to the curve ${{y}^{2}}=4x$ at the point
$\left( A \right)\left( 1,2 \right),\left( B \right)\left( 2,1 \right),\left( C \right)\left( 1,2 \right),\left( D \right)\left( 1,2 \right)$
Ans: The given curve has the following equation:
${{y}^{2}}=4x$.
Differentiating with respect to $x$, we have:
\[2y\frac{dy}{dx}=4\Rightarrow \frac{dy}{dx}=\frac{2}{y}\]
Therefore, the slope of the tangent to the given curve at any point
$\left( x,y \right)$is given by,$\frac{dy}{dx}=\frac{2}{y}$
The given line is $y=x+1$ (which is of the form $mx+c$)
Slope of the line is $1.$
The line $y=x+1$ is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve.
Thus, we must have:
$ \frac{2}{y}=1 $
$ \Rightarrow y=2 $
Now,
$ y=x+1\Rightarrow x=y1 $
$ \Rightarrow x=21=1 $
Hence, the line $y=x+1$ is a tangent to the given curve at the point$\left( 1,2 \right)$
The correct option is A.
Chapter 6 Exercise 6.3 Class 12 Mathematics Solutions
Chapter 6 of class 12 mathematics covers mainly two parts that are tangents and normals; and also the exercise. Math is a very essential subject for students who are studying as it is required for students who have taken up science or plan on doing engineering in the future. It is important for students to understand 12th math as it sets the basis for any hardcore science subjects students plans to take or any engineering course.
Tangents and normals  in chapter 6 exercise 6.3 an overview of the tangent line and the normal line is provided to the students. Here in this chapter theoretical knowledge is given which is very crucial for students to make thorough as they will come to understand all the terms and figures and be able to solve complex problems. It is vital that students understand all the theoretical concepts before moving forward with the subject. In this chapter the equation of a straight line and also the slope at any required point. So within the theory, the slope and the equation and also the slope of the tangent to the curve and its process as a whole is explained. The slope of the normal can be calculated because normal is perpendicular to the tangent. The formulae is generated using this derivation. So by solving all the different equations several functions can be given to understand the multiple functions of the formulae.
Exercise in this chapter 6 exercise 6.3 there are totally 27 questions for students to solve from which are based on the concepts of tangents and normal. Most of the questions are asking students to find the value of the slope based on the tangents of the respective equations. In the questions provided the y axis equations are provided along with its xaxis coordinates. In the sums first, the values of the slopes are calculated and then the coordinates are put into place. This is the best way to solve the sums. Here in these NCERT solutions students will find the step by step method to solve each sum as each problem is different from the other.
Some of the questions that are laid out are to figure out the points where the tangent is parallel to the chord joining the two points. This problem is quite different and more complex than other problems found in chapter 6 exercise 6.3 but the solutions provided here show the step by step method to solve each of them. As you move further in the exercise the problem becomes a little more complex as some of them ask you to form the equation of various slopes.
The calculation of slopes and tangents have been covered in many different ways in Vedantu and explained in ways which can be interpreted easily by students.
Application of Derivatives Exercise 6.3 is based on the following topics:
Tangents and Normals
The lines associated with curves like a circle, parabola, ellipse, and hyperbola are known as tangents and normals. A tangent is a line that touches the curve at a single point, which is known as the point of contact. At the point of contact, the normal is a line perpendicular to the tangent. The normal also passes through the curve's focus.
There are many different tangents that can be drawn to a curve at each point on the curve. As tangents and normals are straight lines, they are written as a linear equation having x and y coordinates. The general version of the tangent and normal equation is ax + by + c = 0. The equation of the tangent and the equation of the curve both satisfy the point of contact.
Finding Tangents And Normals
The lines associated with curves are tangents and normals. Each point on the curve has a tangent, which is a line that touches the curve at a particular location. A line perpendicular to the tangent at the point of contact is called normal.
The equation of the tangent at (x1, y1) is of the form (y  y_{1}) = m(x  x_{1})
The equation of a normal passing through the same point is (y  y_{1}) = 1/m. (x  x_{1}).
Benefits of Chapter 6 Exercise 6.3 Class 12 Mathematics Solved Solutions
These solutions help students to understand the various problems that are underlying in tangents and slopes and how easy to solve them. There are various benefits of using these solutions. The benefits of using these solutions are as follows
The solutions provide the step by step method to solve each problem thereby making it easier for students to understand.
The solutions are solved by experts in the field and students can crosscheck their answers with these solutions to see if they got the right answers.
These solutions are solved according to updated CBSE guidelines so students don't have to worry if what they are doing is relevant or not.
These solutions are useful during the lastminute study as revision material.
Access other exercise solutions of Class 12 Maths Chapter 6  Application of Derivatives
Chapter 6  Application of Derivatives all Exercises in PDF Format  
18 Questions & Solutions  
19 Questions & Solutions  
9 Questions & Solutions  
28 Questions & Solutions 
FAQs on NCERT Solutions for Class 12 Maths Chapter 6: Application of Derivatives  Exercise 6.3
1. How do NCERT solutions help students in class 12?
Class 12 is one of the most crucial chapters of a student's school years as it is the bridge between their school and college, and their grades reflect their performance in school. These NCERT solutions provide students with the best answers and the updated steps to solve them according to the CBSE guidelines. These solutions show students how to solve their sums and students can use the same method that is used here in the exam as well.
2. What does the application of derivatives entail?
The application of derivatives provides students with solving equations based on slopes and tangents. The coordinates of the slopes are provided in some cases and an equation is to be formed based on that. It is important the students understand the theoretical concepts in this chapter only then will they be able to solve the sum further on. Chapter 6 exercise 6.3 contains tangents and normals; and the exercise.
3. What are the topics covered in NCERT solutions for class 12 Maths Chapter 6 Application of Derivatives?
Class 12 Maths Chapter 6 Derivatives has a total of six topics as well as miscellaneous questions. The topics talked about in this chapter are as follows:
Introduction to derivatives
Rate of Change of Quantities
Increase and Decrease of Functions
Tangents and Normals
Approximations
Maxima and Minima
4. How many questions are included in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3?
In the NCERT Solutions for Class 12 Maths Chapter 6 exercise 6.3, a total of 27 have been included with solutions for the ease of understanding of the students. You can download the NCERT Solutions from the Vedantu website or from the Vedantu app at free of cost to ensure clear knowledge and therefore good marks in the exam. The solutions are created by the experts, which means they are a reliable source and offer a wholesome learning experience.
5. What are the mathematical applications of derivatives?
Derivatives are used for a number of important applications in Maths. For example:
Rate of Change of Quantity
Increasing and Decreasing Functions
Tangents and Normals
Approximations
Minimum and Maximum Values
Newton’s Method
6. What are the important formulas in Chapter 6 exercise 6.3 of Class 12 Maths?
Exercise 6.3 has questions based on tangents and normals. It also demands some basic idea of Straight Lines from coordinate geometry. Maximum questions are from the comparison of a slope of the tangent of a curve and a given straight line. The following two equations are important in this regard:
Equation of Tangent
(dy/dx)x = x_{1} ; y = y_{1}
Equation of Normal
A = 1/ (dy/dx)x = x_{1} ; y = y_{1}
7. What are the realtime applications of Derivatives as taught in Class 12?
Derivatives have a lot of important reallife applications. Some of them are:
Calculation of the profit or loss in a business with graphs.
Tracking the variation of temperature.
Determination of speed and distance covered.
In Seismology for finding and tracking the magnitudes of earthquakes.