# 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.