Chain Rule

  • Thread starter jamesbob
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  • #1
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Question: Let
[tex] Q = \sqrt{x^2 + y}e^t[/tex]​
where (for t > or = 0)
[tex] x = \sqrt{1 - e^{-2t}} [/tex]​
and
[tex] y = 2 - e^{-2t} [/tex]​

Using the chain rule calculate dQ/dt, expressing your answer in as simple a form as possible.


My work so far

Subbing in values of x and y:

[tex] Q = \sqrt{1 - e^{-2t} + 2 - e^{-2t}}e^t = \sqrt{3 - 2e^{-2t}}e^t [/tex]

Now applying the product and chain rule:

[tex] u = (3 - 2e^{-2t})^{\frac{1}{2}} \left chain \left rule \Rightarrow u = z^{\frac{1}{2}} \left z = 3 - 2e^-2t [/tex]

[tex] \frac{du}{dz} = \frac{1}{2\sqrt{z}}, \left \frac{dz}{dt} = 4e^-2t [/tex]

So [tex] \frac{du}{dt} = \frac{4e^{-2t}}{2\sqrt{3 - 2e^{-2t}}} [/tex]

As for v, [tex] v = e^t, \left \frac{dv}{dt} = e^t [/tex]

So [tex] \frac{dQ}{dt} = \sqrt{3 - 2e^{-2t}}e^t + \frac{4e^{-t}}{2\sqrt{3 - 2e^{-2t}}} [/tex]

Is this right so far and can it be simplified further?
 

Answers and Replies

  • #2
335
4
jamesbob said:
Question: Let
[tex] Q = \sqrt{x^2 + y}e^t[/tex]​
where (for t > or = 0)
[tex] x = \sqrt{1 - e^{-2t}} [/tex]​
and
[tex] y = 2 - e^{-2t} [/tex]​

Using the chain rule calculate dQ/dt, expressing your answer in as simple a form as possible.


My work so far

Subbing in values of x and y:

[tex] Q = \sqrt{1 - e^{-2t} + 2 - e^{-2t}}e^t = \sqrt{3 - 2e^{-2t}}e^t [/tex]

Now applying the product and chain rule:

[tex] u = (3 - 2e^{-2t})^{\frac{1}{2}} \left chain \left rule \Rightarrow u = z^{\frac{1}{2}} \left z = 3 - 2e^-2t [/tex]

[tex] \frac{du}{dz} = \frac{1}{2\sqrt{z}}, \left \frac{dz}{dt} = 4e^-2t [/tex]

So [tex] \frac{du}{dt} = \frac{4e^{-2t}}{2\sqrt{3 - 2e^{-2t}}} [/tex]

As for v, [tex] v = e^t, \left \frac{dv}{dt} = e^t [/tex]

So [tex] \frac{dQ}{dt} = \sqrt{3 - 2e^{-2t}}e^t + \frac{4e^{-t}}{2\sqrt{3 - 2e^{-2t}}} [/tex]

Is this right so far and can it be simplified further?

Looks good to me. Theoretically you would want to remove the square root from the denominator and add the two expressions, but I personally don't see that that actually simplifies anything.

-Dan
 
  • #3
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Thanks very much :smile:
 
  • #4
Hurkyl
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Subbing in values of x and y:
While a perfectly valid technique, I think part of the point of this problem is not to do such a substitution.
 
  • #5
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Hmm, i see, so would i just do the chian rule keeping x and y in, and then sub values in right at the end?
 
  • #6
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
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Or maybe not even sub in the values at all -- the final result might (or might not) look nicer in terms of x and y!
 

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