Partial differentiation question rocket trajectory

  • #1

Homework Statement


The problem and my attempt are attached

Homework Equations



Chain rule for partial differentiation perhaps

And basic algebra

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
Can you try and differentiate the square root [itex] \sqrt{x(t)^2 + y(t)^2 + z(t)^2} [/itex] to [itex] x(t)^2 + y(t)^2 + z(t)^2 [/itex] first and work from there?
 
  • #3

Homework Statement


The problem and my attempt are attached

Homework Equations



Chain rule for partial differentiation perhaps
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.
AwfulPhysicist said:
And basic algebra

The Attempt at a Solution


I'm unsure of how to approach this but I differentiated all the expression at the top.

In your work it looks like you're trying to integrate both sides. Try to resist that urge.
 
  • #4
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.
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
I've differentiated it using the chain rule, looks horrible

But now I need to get rid of the t's
 
  • #6
hi

How can I differentiate this?! Is it just
(Terms)^1/2 and differentiate using chain rule?
Yes.
AwfulPhysicist said:
Also how would I be able to get rid of the t's at the end?
You don't get rid of them.
 
  • #7
I've differentiated it using the chain rule, looks horrible

But now I need to get rid of the t's
Please post your work here in the form, not as an image, and especially not as an image that is turned on its side.
 
  • #9
Please post your work here in the form, not as an image, and especially not as an image that is turned on its side.
Yeah sorry, sure.
 
  • #10
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?
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
Also how would I be able to get rid of the t's at the end?

[itex] \frac{dr}{dt} [/itex] is a function of [itex] t [/itex] 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.
 
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