Partial Derivative

**JonathanEyoon** · October 30th 2008, 07:56 PM

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.

**U-God** · October 30th 2008, 09:17 PM

Edit nevermind, I got it.

$w = xy$

$\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt}$

$\frac{\partial w}{\partial x} = y$

$\frac{dx}{dt} = -sin(t)$

$\frac{\partial w}{\partial y} = x$

$\frac{dy}{dt} = cos(t)$

sub values in

**JonathanEyoon** · October 31st 2008, 08:29 AM

What do you mean? How does that net me the final answer of cos2t?

**U-God** · October 31st 2008, 03:13 PM

Well, by subbing in those values, we get: $\frac{dw}{dt} = -ysin(t) + xcos(t)$

you defined earlier that $y = sin(t)$ and $x = cos(t)$

so by subbing these to values into $\frac{dw}{dt} = -ysin(t) + xcos(t)$

we get:

$\frac{dw}{dt} = -sin^2(t) + cos^2(t)$

using trigonometric identity $sin^2(t) + cos^2(t) = 1$

we can see that $-sin^2(t) + cos^2(t) = -(1 - cos^2(t)) + cos^2(t) = 2cos^2(t) - 1$

and finally we can use trig identity: $2cos^2(t) - 1 = cos(2t)$

Hope the more elaborated explanation makes more sense,

Thread: Partial Derivative

LinkBack

Thread Tools

Display

Partial Derivative

Similar Math Help Forum Discussions

Derivative of arctan in a partial derivative

how do you take this partial derivative

Partial derivative

partial derivative

partial derivative

Search Tags