Can someone show me how to get the final answer of cos2t for
Find the derivative of w = xy with respect to t along the path x = cost and y = sint.
I know it's chain rule for partials but i'm still confused on how to obtain that final answer.
Can someone show me how to get the final answer of cos2t for
Find the derivative of w = xy with respect to t along the path x = cost and y = sint.
I know it's chain rule for partials but i'm still confused on how to obtain that final answer.
Edit nevermind, I got it.
$\displaystyle w = xy $
$\displaystyle \frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} $
$\displaystyle \frac{\partial w}{\partial x} = y $
$\displaystyle \frac{dx}{dt} = -sin(t) $
$\displaystyle \frac{\partial w}{\partial y} = x $
$\displaystyle \frac{dy}{dt} = cos(t) $
sub values in
Well, by subbing in those values, we get: $\displaystyle \frac{dw}{dt} = -ysin(t) + xcos(t) $
you defined earlier that $\displaystyle y = sin(t) $ and $\displaystyle x = cos(t) $
so by subbing these to values into $\displaystyle \frac{dw}{dt} = -ysin(t) + xcos(t) $
we get:
$\displaystyle \frac{dw}{dt} = -sin^2(t) + cos^2(t) $
using trigonometric identity $\displaystyle sin^2(t) + cos^2(t) = 1 $
we can see that $\displaystyle -sin^2(t) + cos^2(t) = -(1 - cos^2(t)) + cos^2(t) = 2cos^2(t) - 1 $
and finally we can use trig identity: $\displaystyle 2cos^2(t) - 1 = cos(2t) $
Hope the more elaborated explanation makes more sense,