Partial differentiation of cos (in vector calculus)

In summary, the conversation discusses the definition of a sphere in standard spherical polar coordinates and the calculation of the partial differential of Cos(t) with respect to s. The correctness of this calculation is questioned, as it appears to be incorrect based on the equations and general understanding of spherical polar coordinates. The possibility of a trigonometric identity or deliberate mistake is also mentioned.
  • #1
Eight
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Homework Statement



So using standard spherical polar co-ordinates, my notes define a sphere as

r(s,t) = aCos(s)Sin(t) i + aSin(s)Sin(t) j + aCos(t) k

and the normal to the surface is given by the cross product of the two partial differentials:

[itex]\partial[/itex]r[itex]/\partial[/itex]s X [itex]\partial[/itex]r[itex]/dt[/itex]

So my issue is what is in my notes where:

[itex]\partial[/itex]r[itex]/\partial s[/itex]= -aSin(s)Sin(t) i + aCos(s)Sin(t) j + aCot(t) k

It is the final part that I don't understand. Why is the partial differential of Cos(t) with respect to s Cot(t)? I would have calculated it as zero.

Homework Equations



[itex]\partial[/itex][itex]/[/itex][itex]\partial[/itex]s (aCos(t)) = aCot(t) ?

The Attempt at a Solution



All I can think is that it's using a trig identity? As I understand it the k component should only depend upon t, not just from the above equations but in spherical polar co-ordinates in general, at least how I'm picturing it.



Thanks in advance for any help.
 
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  • #2
I agree with you. That looks completely wrong. It should be 0. And it's such a weird mistake that I kept staring at it for some time, wondering if I had misunderstood something. I guess that's what you're doing too. :smile:
 
  • #3
I keep thinking it must be a mistake too, but if it is a mistake then it is deliberate, as it has been repeated in a number of places. And during lectures where we go through the notes it was never picked up on.
 

1. What is partial differentiation of cos?

Partial differentiation of cos is a mathematical technique used in vector calculus to find the rate of change of a function with respect to one of its variables while holding all other variables constant. It is a way to calculate how much a function changes in response to small changes in one of its independent variables.

2. How is partial differentiation of cos different from regular differentiation?

Partial differentiation of cos is different from regular differentiation in that it involves taking the derivative of a multivariable function with respect to one of its variables, while holding all other variables constant. In regular differentiation, we take the derivative of a single-variable function with respect to its input variable.

3. What is the purpose of using partial differentiation in vector calculus?

The purpose of using partial differentiation in vector calculus is to analyze how much a function changes in response to changes in its independent variables. This is useful in many fields of science, such as physics, engineering, and economics, where functions often depend on multiple variables.

4. Can you give an example of how partial differentiation of cos is used in real life?

One example of how partial differentiation of cos is used in real life is in economics. The concept of marginal cost, which measures how much a company's total cost changes when it produces one additional unit of a product, involves taking partial derivatives of the cost function with respect to different input variables, such as labor and materials costs.

5. Are there any common mistakes when performing partial differentiation of cos?

Some common mistakes when performing partial differentiation of cos include forgetting to hold other variables constant, using the chain rule incorrectly, and not properly simplifying the resulting expression. It is important to carefully follow the rules of partial differentiation and double check your work to avoid these errors.

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