Question

x) The slope of the normal to the curve y=sin^2x at π/4 is:
a) 1
b) - 1
c)0
d) 4​

Answer 1

Step-by-step explanation:

Given, y=2x

2

+3sinx

slope =

dx

dy

=4x+3cosx

dy

dy

∣

x=0

=4(0)=3

slope of normal = −1/

dx

dy

=−1/3

Answer 2

Given,

$\[y = {\sin ^2}x\]$

Solution,

Slope of curve,

$\[\frac{{dy}}{{dx}} = 2\sin x\cos x\]$

$\[ \Rightarrow {\frac{{dy}}{{dx}}_{\left( {x = \frac{\pi }{4}} \right)}} = \sin 2x\]$

$\[ \Rightarrow {\frac{{dy}}{{dx}}_{\left( {x = \frac{\pi }{4}} \right)}} = \sin 2 \times \left( {\frac{\pi }{4}} \right)\]$

$\[ \Rightarrow {\frac{{dy}}{{dx}}_{\left( {x = \frac{\pi }{4}} \right)}} = \sin \left( {\frac{\pi }{2}} \right)\]$

$\[ \Rightarrow {\frac{{dy}}{{dx}}_{\left( {x = \frac{\pi }{4}} \right)}} = 1\]$

Slope of the normal to the curve,

$\[ = - \frac{1}{{\left( {\frac{{dy}}{{dx}}} \right)}}\,\]$

$\[ = \frac{{ - 1}}{1}\]$

$\[ = - 1\]$

Hence the slope of normal to the curve is -1.

The correct option is (b), i.e. -1.