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Chapter 15: Problem 2

Calculate the rate of change of \(i(t)=4 \sin 2 t+3 t\) when (a) \(t=\frac{\pi}{3}\) (b) \(t=0.6\).

Short Answer

Expert verified
Answer: (a) The rate of change at \(t = \frac{\pi}{3}\) is -1. (b) The rate of change at \(t = 0.6\) is approximately 2.9.

Step by step solution

01

Find the derivative of i(t)

The given function is \(i(t)=4 \sin 2t+3t\). We'll find its derivative with respect to t using basic differentiation rules. The derivative of \(\sin(2t)\) is \(2\cos(2t)\), and the derivative of \(3t\) is \(3\). Therefore, the derivative of i(t) is given by: \(i'(t) = 4(2\cos(2t)) + 3\) Simplifying, we get: \(i'(t) = 8\cos(2t) + 3\)
02

Calculate i'(t) when t = π/3

In this step, we'll substitute \(t=\frac{\pi}{3}\) into the expression for i'(t) and find the rate of change at this point: \(i'\left(\frac{\pi}{3}\right) = 8\cos\left(2\cdot\frac{\pi}{3}\right) + 3\) Since \(\cos(2\cdot\frac{\pi}{3}) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\), the expression becomes: \(i'\left(\frac{\pi}{3}\right) = 8\left(-\frac{1}{2}\right) + 3 = -4 + 3 = -1\) So, the rate of change of i(t) when \(t=\frac{\pi}{3}\) is -1.
03

Calculate i'(t) when t = 0.6

Now, we'll substitute \(t=0.6\) into the expression for i'(t) and find the rate of change at this point: \(i'(0.6) = 8\cos(2 \cdot 0.6) + 3\) Using a calculator or software to find the value of \(\cos(1.2)\), we get: \(i'(0.6) \approx 8 \cdot 0.362357754 + 3 \approx 2.898862028\) So, the rate of change of i(t) when \(t=0.6\) is approximately 2.9
04

Conclusion

The rate of change of the function i(t) at \(t=\frac{\pi}{3}\) is -1 and at \(t=0.6\) is approximately 2.9.

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Most popular questions from this chapter

Evaluate the rate of change of \(H(t)=5 \sin t-3 \cos 2 t\) when (a) \(t=0\) (b) \(t=1.3\).

Find the rate of change of the following functions: (a) \(\frac{3 t^{3}-t^{2}}{2 t}\) (b) \(\ln \sqrt{x}\) (c) \((t+2)(2 t-1)\) (d) \(\mathrm{e}^{3 v}\left(1-\mathrm{e}^{v}\right)\) (e) \(\sqrt{x}(\sqrt{x}-1)\)

The function \(y(x)\) is given by $$ y(x)=2 x^{3}-9 x^{2}+1 $$ (a) Calculate the values of \(x\) for which \(y^{\prime}=0\). (b) Calculate the values of \(x\) for which \(y^{\prime \prime}=0\). (c) State the interval(s) on which \(y\) is increasing. (d) State the interval(s) on which \(y\) is decreasing. (e) State the interval(s) on which \(y^{\prime}\) is increasing. (f) State the interval(s) on which \(y^{\prime}\) is decreasing.

The function \(y(x)\) is given by \(y(x)=1-\cos x\). Find the values of \(x\) where (a) \(y^{\prime}=0\), (b) \(y^{\prime \prime}=0\).

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