Finding specified partial derivatives

In summary, the homework statement is to solve a equation with given values for r and s. The Attempt at a Solution provides a derivation of the equation, but the problem is that the images for the equation are not loading.
  • #1
dalarev
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0

Homework Statement



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Homework Equations



attachment.php?attachmentid=11325&stc=1&d=1193283695.gif


The Attempt at a Solution



I follow the steps until I get to 2(x+y+z)(1 - sin(r+s) + cos(r+s))

The actual derivation process isn't the problem, I get lost in trying to figure out when to plug values for r and s.
 

Attachments

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  • partialder.gif
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  • #2
Images don't load.
 
  • #3
images loaded fine for me...
anyhow, you have done everything correctly thus far. just plug the given values for s and r into the final expression 2(x+y+z)(1 - sin(r+s) + cos(r+s))

you will never go awry if you always plug in given values last
 
  • #4
paradigm said:
images loaded fine for me...
anyhow, you have done everything correctly thus far. just plug the given values for s and r into the final expression 2(x+y+z)(1 - sin(r+s) + cos(r+s))

I'm guessing this is after I replace x,y,z with their respective functions, correct?

edit: Negative. I did that and got 0 as an answer. According to the answer guide (back of the book), the correct answer is 13.
 
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  • #5
yes, I'm sorry, i should have specified... plug in x,y,z and then s, r
 
  • #6
paradigm said:
yes, I'm sorry, i should have specified... plug in x,y,z and then s, r

That's not correct though, I get 6(0) = 0, correct answer should be 13.
 
  • #7
i'm currently unable to view the images you attached... what were x,y, and z again?

EDIT: i recall r and s being 1 and -1, right? if that's the case (1 - sin(r+s) + cos(r+s) should be equal to (1 - sin0 + cos0) = (1 - 0 +1) = 2, not zero..
 
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  • #8
x = r - s
y = cos(r+s)
z = sin(r+s)
 
  • #9
so if r was 1 and s was -1, then you should have 2(2+1)(2) = 12... but you say the answer was 13?
 
  • #10
paradigm said:
so if r was 1 and s was -1, then you should have 2(2+1)(2) = 12... but you say the answer was 13?

Yeah, that's what I got too. But then again, the answer guide has been known to be wrong on more than a couple of occasions. Disregard further posts, and thanks for your help.
 

FAQ: Finding specified partial derivatives

What is a partial derivative?

A partial derivative is a mathematical concept used to measure the rate of change of a function with respect to one of its variables, while holding all other variables constant. It is denoted by ∂ and is often used in multivariable calculus to solve problems involving functions with multiple independent variables.

How do you find a specified partial derivative?

To find a specified partial derivative, you must first identify the function and the independent variable that you want to differentiate. Then, you differentiate the function with respect to that variable, treating all other variables as constants. This will give you the partial derivative of the function with respect to the specified variable.

Why are partial derivatives important in science?

Partial derivatives are important in science because they allow us to analyze how a function changes in response to changes in its independent variables. This is important in many fields of science, including physics, chemistry, and economics, where functions with multiple variables are commonly used to model real-world phenomena.

What is the difference between a partial derivative and a regular derivative?

A regular derivative measures the rate of change of a function with respect to its independent variable, while a partial derivative measures the rate of change of a function with respect to one of its variables, holding all other variables constant. In other words, a regular derivative considers only one variable, while a partial derivative takes into account the effect of multiple variables on a function.

Can you give an example of how partial derivatives are used in science?

One example of how partial derivatives are used in science is in thermodynamics. The partial derivative of temperature with respect to pressure is used to calculate the isothermal compressibility of a substance, which is an important property in determining its response to changes in pressure. This is just one of many applications of partial derivatives in science.

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