Question

Use the chain rule to find dQ/dt, where Q = sqrt(5x^2 + 5y^2 + 5z^2), x=cost, y=sint, and z=cost. Express in terms of t.

Answer 1

\[Find: \frac{ dQ }{ dt }, where: Q=\sqrt{5x^2 +5y^2 + 5z^2}, x=\cos t, y=\sin t, z=\cos t\]

Answer 2

Recall that the chain rule is \(\dfrac{dQ}{dt} = \dfrac{\partial Q}{\partial x}\dfrac{dx}{dt} + \dfrac{\partial Q}{\partial y}\dfrac{dy}{dt} +\dfrac{\partial Q}{\partial z}\dfrac{dz}{dt}\). Did you find each of those derivatives?

Answer 3

Or are you not in calc 3? XD

Answer 4

Vector Calc

Answer 5

Uncertain about the partails:
\[\frac{ ∂Q }{ ∂x} = \sqrt{10x+5y^2+4z^2} ?\]

Answer 6

... how do you enter the other d symbol?

Answer 7

that was in reference to the first partial

Answer 8

Unfortunately, that's not it. First note that \(\sqrt{5x^2+5y^2+5z^2} = (5x^2+5y^2+5z^2)^{1/2}\).

Therefore,

\[\begin{aligned}\dfrac{\partial Q}{\partial x} &= \frac{1}{2}(5x^2+5y^2+5z^2)^{-1/2}\cdot \frac{\partial}{\partial x}(5x^2+5y^2+5z^2)\\ &= \frac{1}{2\sqrt{5x^2+5y^2+5z^2}}\cdot 10x \\ &= \frac{5x}{\sqrt{5x^2+5y^2+5z^2}} \end{aligned}\]

\(\dfrac{\partial Q}{\partial y}\) and \(\dfrac{\partial Q}{\partial z}\) are found in the same exact way. Does this make sense?

Also, in \(\LaTeX \), \partial gives you \(\partial \).

Answer 9

makes sense

Answer 10

and just to make sure:
dx/dt refers ONLY to x=cost?

Answer 11

Yes, that would be correct. Same thing applies for \(\dfrac{dy}{dt}\) and \(\dfrac{dz}{dt}\).

Answer 12

so it ends up being this?
\[\frac{dQ}{dt} = \frac{ -15\cos t \sin t }{ \sqrt{10\cos^2 t + 5\sin^2 t} }\]

Answer 13

I get \(\dfrac{dQ}{dt} = \dfrac{\color{red}{-5}\cos t\sin t}{\sqrt{10\cos^2 t+5\sin^2t}} = -\dfrac{5\cos t\sin t}{\sqrt{5+5\cos^2 t}}\)

Answer 14

Oh, I see. I somehow subtracted the whole y section and it made it look like all 3 sections were identical.

Answer 15

Thanks for the help