Question

If the normal to the curve \[y = f\left( x \right)\] at the point \[\left( {3,4} \right)\] make an angle $\dfrac{3{\pi }}{4}$ with the positive \[x\] -axis, then \[f'\left( 3 \right)\] is?

Answer 1

Hint: First, we shall analyze the given information. We are given that the normal to the curve\[y = f\left( x \right)\] is at the point \[\left( {3,4} \right)\] make an angle $\dfrac{3{\pi }}{4}$
We shall use the formula to calculate the slope of normal at the point \[\left( {3,4} \right)\] and also it is given that the normal to the curve makes an angle $\dfrac{3{\pi }}{4}$
We need to equate the resultant equations so that we can solve this problem easily.
Formula to be used:
    The formula to calculate the slope of the normal is as follows.
The slope of the normal $ = \dfrac{{ - 1}}{{f'\left( x \right)}}$
The formula to calculate the angle of a tangent, when two angles are given is as follows.
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$

Complete step by step answer:
It is given that the normal to the curve\[y = f\left( x \right)\] is at the point \[\left( {3,4} \right)\]
The given angle is $\dfrac{3{\pi }}{4}$with the positive \[x\]-axis
We have to find \[f'\left( 3 \right)\] when the angle is on the positive \[x\]-axis.
We know that from hint part slope of the tangent \[y = f\left( x \right)\] is $\dfrac{{dy}}{{dx}} = f'\left( x \right)$
At the point-slope of normal $ = \dfrac{{ - 1}}{{f'\left( 3 \right)}}$
It is given that the normal to the curve makes an angle $\dfrac{3{\pi }}{4}$
Hence, the slope of normal is
$m = \tan \left( {\dfrac{3\pi}{4}} \right)$
We shall equate the obtained equations.
That is$\dfrac{{ - 1}}{{f'\left( 3 \right)}} = \tan \left( {\dfrac{3\pi }{4}} \right)$……..$\left( 1 \right)$
By using the trigonometric function we can find $\tan \left( {\dfrac{3\pi }{4}} \right)$\[ = tan\left( {\pi - \dfrac{\pi }{4}\;\;} \right)\;\]
The formula to calculate the angle of a tangent, when two angles are given is as follows.
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
We need to use the above formula.
\[tan\left( {\pi - \dfrac{\pi }{4}\;\;} \right)\; = \dfrac{{\tan \pi - \tan \dfrac{\pi }{4}}}{{1 + \tan \pi \tan \dfrac{\pi }{4}}}\]
Since$\tan \pi = 0$ and $\tan \dfrac{\pi }{4} = 1$ , we get
\[tan\left( {\pi - \dfrac{\pi }{4}\;\;} \right)\; = \dfrac{{0 - 1}}{{1 + 0 \times 1}}\]
\[ \Rightarrow tan\left( {\pi - \dfrac{\pi }{4}\;\;} \right)\; = - 1\]
Hence, $\tan \left( {\dfrac{3\pi }{4}} \right)$\[ = tan\left( {\pi - \dfrac{\pi }{4}\;\;} \right)\; = - 1\]

$ \Rightarrow \tan \left( {3\dfrac{\pi }{4}} \right) = - 1$ …….$\left( 2 \right)$
Now, we shall substitute the value of the equation$\left( 2 \right)$ in the first equation.
$\dfrac{{ - 1}}{{f'\left( 3 \right)}} = \tan \left( {\dfrac{3\pi }{4}} \right)$
 $ = - 1$
We need to reciprocate on both sides.
That is
\[ - f'\left( 3 \right) = - 1\]
\[ \Rightarrow f'\left( 3 \right) = 1\] [In the above equation minus and minus gets canceled]
 Therefore, \[f'\left( 3 \right) = 1\]

Note: Since it is given that the normal to the curve \[y = f\left( x \right)\] is at the point \[\left( {3,4} \right)\] make an angle $\dfrac{3{\pi }}{4}$, we compared the slope of the normal to the curve \[y = f\left( x \right)\] at the given point with the slope of the normal for the given angle.