Partial differentiation question rocket trajectory

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Homework Help Overview

The discussion revolves around a problem related to partial differentiation in the context of rocket trajectory, specifically focusing on differentiating expressions involving the coordinates x(t), y(t), and z(t) as functions of time t.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss differentiating the expression for the distance involving the square root of the sum of squares of the coordinates. There are attempts to apply the chain rule and questions about how to handle the variable t in the differentiation process.

Discussion Status

Several participants have shared their attempts at differentiation and expressed uncertainty about the approach. Guidance has been offered regarding the application of the chain rule and the nature of the resulting expressions. There is an ongoing exploration of how to express the results in terms of constants and the role of the variable t.

Contextual Notes

Some participants note the expectation of partial differentiation in the problem, raising questions about the relevance of the approach being discussed. There is also mention of homework constraints regarding the format of submissions.

AwfulPhysicist
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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.
 

Attachments

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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?
 
AwfulPhysicist said:

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.
 
Mark44 said:
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?
 

Attachments

  • image.jpg
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I've differentiated it using the chain rule, looks horrible

But now I need to get rid of the t's
 
AwfulPhysicist said:
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.
 
AwfulPhysicist said:
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.
 

Attachments

  • image.jpg
    image.jpg
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Mark44 said:
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
AwfulPhysicist said:
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
AwfulPhysicist said:
Also how would I be able to get rid of the t's at the end?

\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.
 

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