Question

Find the equation of the tangent to the curve.

y=xsinx when x= pi

I am so far at

y=-cospi(x) +b

But I need to get b, and I'm stuck.

Isn't the y value just xsinx
and the x value just pi?

Answer 1

y'=xcosx+sinx

Answer 2

dy/dx=-pi

Answer 3

Yes, sorry. I mean -pi

Answer 4

when x=pi,
y=0

Answer 5

Ah, Ok.

Answer 6

\[y-0=-\pi(x-\pi)\]
\[y=-\pi(x)+\pi^2\]

Answer 7

Yes, thanks very much :)

Answer 8

yw

Answer 9

find the deriviative
f(x)' = sinx + xcosx

sub x = pi

\[f(\pi) = \sin(\pi) + \pi \times \cos(\pi) \]

sin( pi) = 0 cos(pi) = -1

then the gradient at x = pi is m = -pi

to find the y value in the point sub x = pi into the equation

y = pi x sin(pi)

y = 0

use point gadient formula

y - 0 = -pi( x - pi)