jamesbob
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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:
Q = \sqrt{1 - e^{-2t} + 2 - e^{-2t}}e^t = \sqrt{3 - 2e^{-2t}}e^t
Now applying the product and chain rule:
u = (3 - 2e^{-2t})^{\frac{1}{2}} \left chain \left rule \Rightarrow u = z^{\frac{1}{2}} \left z = 3 - 2e^-2t
\frac{du}{dz} = \frac{1}{2\sqrt{z}}, \left \frac{dz}{dt} = 4e^-2t
So \frac{du}{dt} = \frac{4e^{-2t}}{2\sqrt{3 - 2e^{-2t}}}
As for v, v = e^t, \left \frac{dv}{dt} = e^t
So \frac{dQ}{dt} = \sqrt{3 - 2e^{-2t}}e^t + \frac{4e^{-t}}{2\sqrt{3 - 2e^{-2t}}}
Is this right so far and can it be simplified further?
Q = \sqrt{x^2 + y}e^t
where (for t > or = 0)x = \sqrt{1 - e^{-2t}}
andy = 2 - e^{-2t}
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:
Q = \sqrt{1 - e^{-2t} + 2 - e^{-2t}}e^t = \sqrt{3 - 2e^{-2t}}e^t
Now applying the product and chain rule:
u = (3 - 2e^{-2t})^{\frac{1}{2}} \left chain \left rule \Rightarrow u = z^{\frac{1}{2}} \left z = 3 - 2e^-2t
\frac{du}{dz} = \frac{1}{2\sqrt{z}}, \left \frac{dz}{dt} = 4e^-2t
So \frac{du}{dt} = \frac{4e^{-2t}}{2\sqrt{3 - 2e^{-2t}}}
As for v, v = e^t, \left \frac{dv}{dt} = e^t
So \frac{dQ}{dt} = \sqrt{3 - 2e^{-2t}}e^t + \frac{4e^{-t}}{2\sqrt{3 - 2e^{-2t}}}
Is this right so far and can it be simplified further?