- #1
jamesbob
- 63
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Question: Let
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?
[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?