Question

How do you find the equation of the tangent to the curve \[x={{t}^{4}}+1\], \[y={{t}^{3}}+t\] at the point where \[t=-1\]?

Answer 1

Hint: This question is from the topic of calculus. In solving this question, we will differentiate the equation \[x={{t}^{4}}+1\] with respect to t. After that, we will differentiate the equation \[y={{t}^{3}}+t\] with respect to t. After that, we will find out the value of \[\dfrac{dy}{dx}\]. In the equation of \[\dfrac{dy}{dx}\], we will put the value of t as -1. After that, we will use the formula of straight lines and get the answer.

Complete step-by-step solution:
Let us solve this question.
In this question, we have asked to find the equation of tangent of the curve. The curve is given as \[x={{t}^{4}}+1\] and \[y={{t}^{3}}+t\]. We have to find the tangent at t=-1.
Let us first differentiate the equation \[x={{t}^{4}}+1\] with respect to t, we can write
\[\dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{t}^{4}}+1 \right)\]
Using the addition rule of differentiation: \[\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)\dfrac{d}{dx}\left( g\left( x \right) \right)\] [where f(x) and g(x) are two function of x], we can write
\[\dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{t}^{4}} \right)+\dfrac{d}{dt}\left( 1 \right)\]
As we know that differentiation of any constant is always zero, so we can write
\[\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{t}^{4}} \right)+0\]
Using the formula of differentiation: \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\], we can write
\[\Rightarrow \dfrac{dx}{dt}=4{{t}^{4-1}}\]
\[\Rightarrow \dfrac{dx}{dt}=4{{t}^{3}}\]
Now, we will differentiate the equation \[y={{t}^{3}}+t\] with respect to t, we can write
\[\dfrac{dy}{dt}=\dfrac{d}{dt}\left( {{t}^{3}}+t \right)\]
\[\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( {{t}^{3}}+{{t}^{1}} \right)\]
\[\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( {{t}^{3}} \right)+\dfrac{d}{dt}\left( {{t}^{1}} \right)\]
Using the formula: \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\], we can write
\[\Rightarrow \dfrac{dy}{dt}=3{{t}^{3-1}}+1\]
\[\Rightarrow \dfrac{dy}{dt}=3{{t}^{2}}+1\]
As we know that the tangent of any curve is always equal to \[\dfrac{dy}{dx}\], so we will divide the equation \[\dfrac{dy}{dt}=3{{t}^{2}}+1\] by the equation \[\dfrac{dx}{dt}=4{{t}^{3}}\], we can write
\[\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3{{t}^{2}}+1}{4{{t}^{3}}}\]
The above equation can also be written as
\[\Rightarrow \dfrac{dy}{dt}\times \dfrac{dt}{dx}=\dfrac{3{{t}^{2}}+1}{4{{t}^{3}}}\]
The above equation can also be written as
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{3{{t}^{2}}+1}{4{{t}^{3}}}\]
Now, we will put here the value of t as -1, we will get
\[\Rightarrow {{\left. \dfrac{dy}{dx} \right|}_{t=-1}}=\dfrac{3{{\left( -1 \right)}^{2}}+1}{4{{\left( -1 \right)}^{3}}}\]
The above equation can also be written as
\[\Rightarrow {{\left. \dfrac{dy}{dx} \right|}_{t=-1}}=\dfrac{3+1}{4\left( -1 \right)}=\dfrac{4}{-4}\]
The above equation can also be written as
\[\Rightarrow {{\left. \dfrac{dy}{dx} \right|}_{t=-1}}=-1\]
Hence, we get the slope at t=-1 as -1.
Now, we will find the value of x and y at t=-1.
So, at t=-1, the value of x will be
\[x={{\left( -1 \right)}^{4}}+1=1+1=2\]
And, at t=-1, the value of y will be
\[y={{\left( -1 \right)}^{3}}+\left( -1 \right)=-1-1=-2\]
Now, using the co-ordinate \[\left( {{x}_{1}},{{y}_{1}} \right)\] as (2,-2) and slope as -1, we can write the straight line as
\[\left( y-{{y}_{1}} \right)=\left( \dfrac{dy}{dx} \right)\left( x-{{x}_{1}} \right)\]
\[\Rightarrow \left( y-\left( -2 \right) \right)=\left( -1 \right)\left( x-\left( 2 \right) \right)\]
The above can also be written as
\[\Rightarrow y+2=\left( -1 \right)\left( x-2 \right)\]
\[\Rightarrow y+2=-x+2\]
\[\Rightarrow x+y+2-2=0\]
The above equation can also be written as
\[\Rightarrow x+y=0\]
Hence, the equation of tangent at t=-1 is \[x+y=0\].

Note: We should have a better knowledge in the topic of calculus to solve this type of question easily. We should remember the following formulas of differentiation:
Addition rule of differentiation: \[\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)\dfrac{d}{dx}\left( g\left( x \right) \right)\]
\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]
We should know that if a point is given as \[\left( {{x}_{1}},{{y}_{1}} \right)\] and the slope is \[\dfrac{dy}{dx}\], then the equation of line will be
\[\left( y-{{y}_{1}} \right)=\left( \dfrac{dy}{dx} \right)\left( x-{{x}_{1}} \right)\]