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Chapter 16: Problem 3

Find the rate of change of $$ q(t)=2 \mathrm{e}^{-t / 2} \cos 2 t $$ when \(t=1\)

Short Answer

Expert verified
Question: Find the rate of change of the function \(q(t)=2e^{-t/2}\cos{2t}\) when \(t=1\). Answer: To find the rate of change of the function \(q(t)\) at \(t=1\), we calculated the derivative \(q'(t)\) using both the product rule and the chain rule. We found that \(q'(t) = -e^{-t/2}\cos{2t} + 2e^{-t/2}(-2\sin{2t})\). When \(t=1\), the rate of change is \(q'(1) = -e^{-1/2}\cos{2} + 2e^{-1/2}(-2\sin{2})\).

Step by step solution

01

Identify the product and chain rules

The given function \(q(t)=2e^{-t/2}\cos{2t}\) involves a product of two functions of t and each function also involves the composition of functions. This indicates that both the product rule and the chain rule will be required while differentiating q(t).
02

Apply product rule and chain rule

We'll first apply the product rule which is \((fg)'=f'g+fg'\). Let \(f(t)=2e^{-t/2}\) and \(g(t)=\cos{2t}\). Now, we need to find the derivatives of \(f(t)\) and \(g(t)\). For \(f(t)\), we have the chain rule which is \((h(k(t)))'=h'(k(t))k'(t)\). Let \(h(t)=2e^t\) and \(k(t)=-t/2\). Then $$ f'(t) = (2e^{-t/2})' = 2e^{-t/2}(-\frac{1}{2}) = -e^{-t/2}. $$ Now, differentiate \(g(t)=\cos{2t}\): $$ g'(t) = (-2\sin{2t}). $$
03

Combine derivatives

Now, we will substitute \(f'(t)\) and \(g'(t)\) into the product rule formula to find the derivative of \(q(t)\) with respect to \(t\). $$ q'(t) = f'(t)g(t)+ f(t)g'(t) = -e^{-t/2}\cos{2t} + 2e^{-t/2}(-2\sin{2t}). $$
04

Evaluate q'(t) at t=1

Now let's evaluate \(q'(t)\) when \(t=1\): $$ q'(1) = -e^{-1/2}\cos{2} + 2e^{-1/2}(-2\sin{2}). $$ This gives us the rate of change of \(q(t)\) at \(t=1\).

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