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Question: If the normal to the curve \[y = f\left( x \right)\] at the point \[\left( {3,4} \right)\] make an a...

If the normal to the curve y=f(x)y = f\left( x \right) at the point (3,4)\left( {3,4} \right) make an angle 3π4\dfrac{3{\pi }}{4} with the positive xx -axis, then f′(3)f'\left( 3 \right) is?

Explanation

Solution

First, we shall analyze the given information. We are given that the normal to the curvey=f(x)y = f\left( x \right) is at the point (3,4)\left( {3,4} \right) make an angle 3π4\dfrac{3{\pi }}{4}
We shall use the formula to calculate the slope of normal at the point (3,4)\left( {3,4} \right) and also it is given that the normal to the curve makes an angle 3π4\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 =−1f′(x) = \dfrac{{ - 1}}{{f'\left( x \right)}}
The formula to calculate the angle of a tangent, when two angles are given is as follows.
tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\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 curvey=f(x)y = f\left( x \right) is at the point (3,4)\left( {3,4} \right)
The given angle is 3π4\dfrac{3{\pi }}{4}with the positive xx-axis
We have to find f′(3)f'\left( 3 \right) when the angle is on the positive xx-axis.
We know that from hint part slope of the tangent y=f(x)y = f\left( x \right) is dydx=f′(x)\dfrac{{dy}}{{dx}} = f'\left( x \right)
At the point-slope of normal =−1f′(3) = \dfrac{{ - 1}}{{f'\left( 3 \right)}}
It is given that the normal to the curve makes an angle 3π4\dfrac{3{\pi }}{4}
Hence, the slope of normal is
m=tan⁡(3π4)m = \tan \left( {\dfrac{3\pi}{4}} \right)
We shall equate the obtained equations.
That is−1f′(3)=tan⁡(3π4)\dfrac{{ - 1}}{{f'\left( 3 \right)}} = \tan \left( {\dfrac{3\pi }{4}} \right)……..(1)\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 = 0andand\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(π−π4;;);=−1 = tan\left( {\pi - \dfrac{\pi }{4};;} \right); = - 1

⇒tan⁡(3π4)=−1 \Rightarrow \tan \left( {3\dfrac{\pi }{4}} \right) = - 1 …….(2)\left( 2 \right)
Now, we shall substitute the value of the equation(2)\left( 2 \right) in the first equation.
−1f′(3)=tan⁡(3π4)\dfrac{{ - 1}}{{f'\left( 3 \right)}} = \tan \left( {\dfrac{3\pi }{4}} \right)
=−1= - 1
We need to reciprocate on both sides.
That is
−f′(3)=−1- f'\left( 3 \right) = - 1
⇒f′(3)=1\Rightarrow f'\left( 3 \right) = 1 [In the above equation minus and minus gets canceled]
Therefore, f′(3)=1f'\left( 3 \right) = 1

Note: Since it is given that the normal to the curve y=f(x)y = f\left( x \right) is at the point (3,4)\left( {3,4} \right) make an angle 3π4\dfrac{3{\pi }}{4}, we compared the slope of the normal to the curve y=f(x)y = f\left( x \right) at the given point with the slope of the normal for the given angle.