Question

the equation of normal to the curve y=sin^2x at x=3.14/2

Answer 1

Answer: Given  

y

=

sin

(

2

x

)

at  

x

=

π

2

solve for the point first

y

=

sin

(

2

x

)

y

=

sin

(

2

(

π

2

)

)

y

=

0

Our point  

(

x

1

,

y

1

)

=

(

π

2

,

0

)

Let us solve the slope m of the tangent line

y

=

sin

(

2

x

)

y

'

=

cos

(

2

x

)

⋅

d

d

x

(

2

x

)

=

cos

(

2

x

)

(

2

)

=

2

⋅

cos

(

2

x

)

m

=

2

⋅

cos

(

2

(

π

2

)

)

=

2

⋅

(

−

1

)

=

−

2

m

=

−

2

The Tangent Line

y

−

y

1

=

m

(

x

−

x

1

)

y

−

0

=

−

2

(

x

−

π

2

)

y= -2x +π