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Calculate $\frac{\partial u}{\partial t} \text { where } u = x^2 + y^2 + xz \text{ and } x = \sin t, y = e^t, z = t^3.$

The book claims the answer is $\sin2t + 2e^{2t} + 3t^2\sin t + t^3\cos t$.

I've got $(2 \sin t + t^3) \cos t + 2e^{t^2} + 3t^2 \sin t$.

Where is my mistake?

Using the chain rule, I've got that:

$$u'_t = u_xx_t + u_y y_t + u_zz_t = (2x + z)\cos t + 2y \cdot e^t + x 3t^2$$

Plug $x, y , z$ and finally get:

$\frac{\partial u}{\partial t} = (2\sin t + t^3)\cos t + 2 e^te^t + \sin t 3t^2 = \color{blue}{2\sin t \cos t + t^3 \cos t + 2e^{2t} + 3t^2\sin t}$.

I wonder there is my mistake? thanks in advance.

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  • $\begingroup$ Note that $ \ \sin 2t \ = \ 2 \ \sin t \ \cos t \ ... $ (I tell students that there are three identities from trig that they should be sure to remember, if they forget everything else: the Pythagorean Identity and the double-angle formulas for sine and cosine.) $\endgroup$ – colormegone Feb 14 '14 at 2:26
  • $\begingroup$ @RecklessReckoner So my answer is right? $\endgroup$ – Billie Feb 14 '14 at 2:27
  • $\begingroup$ The rest of what you wrote in blue agrees with the textbook answer. What you wrote the first time is not quite correct because $ \ e^t \cdot e^t \ \neq \ e^{t^2} \ . $ (Oh, and you should drop the prime on $ \ u_t \ , $ since you are already indicating differentiation by the subscript.) $\endgroup$ – colormegone Feb 14 '14 at 2:30
  • $\begingroup$ Thank you very much :) Can you please explain my why $e^t\cdot e^t \ne e^{t^2}? \text{ Isn't } e^{t^2} = e^{t\cdot2} = e^{2t}?$ $\endgroup$ – Billie Feb 14 '14 at 2:40
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    $\begingroup$ Reading your second question again, I believe you are thinking of $ \ (e^t)^2 \ $ , which is $ \ (e^t) \cdot (e^t) \ = \ e^{2t} \ . $ But $ \ e^{t^2} \ $ is taken to mean $ \ e^{(t^2)} \ . $ (So this is a notational issue to be aware of...) $\endgroup$ – colormegone Feb 14 '14 at 2:51

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