Partial differentiation question rocket trajectory

1. Jan 4, 2015

AwfulPhysicist

1. The problem statement, all variables and given/known data
The problem and my attempt are attached

2. Relevant equations

Chain rule for partial differentiation perhaps

And basic algebra
3. The attempt at a solution
I'm unsure of how to approach this but I differentiated all the expression at the top.

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2. Jan 4, 2015

barefeet

Can you try and differentiate the square root $\sqrt{x(t)^2 + y(t)^2 + z(t)^2}$ to $x(t)^2 + y(t)^2 + z(t)^2$ first and work from there?

3. Jan 4, 2015

Staff: Mentor

No. The three coordinate functions x(t), y(t), and z(t) are functions of t alone, and you're asked to find dr/dt.

Calculate x2 + y2 + z2, then take the square root, then differentiate with respect to t.
In your work it looks like you're trying to integrate both sides. Try to resist that urge.

4. Jan 4, 2015

AwfulPhysicist

hi

How can I differentiate this?! Is it just
(Terms)^1/2 and differentiate using chain rule?

Also how would I be able to get rid of the t's at the end?

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5. Jan 4, 2015

AwfulPhysicist

I've differentiated it using the chain rule, looks horrible

But now I need to get rid of the t's

6. Jan 4, 2015

Staff: Mentor

Yes.
You don't get rid of them.

7. Jan 4, 2015

Staff: Mentor

Please post your work here in the form, not as an image, and especially not as an image that is turned on its side.

8. Jan 4, 2015

AwfulPhysicist

The question wanted it in terms of the constants, the topic I'm doing involves partial differentiation it seems weird that there is no partial differentiation

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9. Jan 4, 2015

AwfulPhysicist

Yeah sorry, sure.

10. Jan 4, 2015

SammyS

Staff Emeritus
Yes, it is just (Terms)1/2 and differentiate using chain rule .

You don't get rid of the time. The resulting expression for speed is a function of time, t .

11. Jan 4, 2015

Stephen Tashi

$\frac{dr}{dt}$ is a function of $t$ so you don't want to get rid of the t's. The t's might cancel out in some problems, but, in general, we expect a function of t to have t's in it.