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Chapter 11: Problem 39

Compute the following derivatives. $$\frac{d}{d t}\left(\left(3 t^{2} \mathbf{i}+\sqrt{t} \mathbf{j}-2 t^{-1} \mathbf{k}\right) \cdot(\cos t \mathbf{i}+\sin 2 t \mathbf{j}-3 t \mathbf{k})\right)$$

Short Answer

Expert verified
Question: Compute the time derivative of the vector function \( (3t^2 \mathbf{i}+\sqrt{t} \mathbf{j}-2t^{-1} \mathbf{k}) \cdot(\cos t \mathbf{i}+\sin 2t \mathbf{j}-3t \mathbf{k}) \). Answer: The time derivative of the given vector function is \( (6t\cos t - 3t^2\sin t) \mathbf{i} + (\frac{1}{2\sqrt{t}}(\sin 2t) + \sqrt{t}(2\cos 2t)) \mathbf{j} + (6(-2t^{-2}(-3t)) + (-2t^{-1})(-3)) \mathbf{k} \).

Step by step solution

01

Derivative of the first component

First, we compute the derivative of the component in the direction of \(\mathbf{i}\), which is given by the product of \(3t^2\) and \(\cos t\). We use the product rule: $$(fg)' = f'g + fg'$$ So for our case, we have $$\frac{d}{dt}(3t^2\cos t) = (6t\cos t) - (3t^2\sin t)$$
02

Derivative of the second component

Now, we compute the derivative of the component in the direction of \(\mathbf{j}\), which is given by the product of \(\sqrt{t}\) and \(\sin 2t\). We use the product rule and the chain rule for the square root function: $$(\sqrt{t})' = \frac{1}{2\sqrt{t}}$$ So for our case, we have $$\frac{d}{dt}(\sqrt{t}\sin 2t) = \frac{1}{2\sqrt{t}}(\sin 2t) + \sqrt{t}(2\cos 2t)$$
03

Derivative of the third component

Finally, we compute the derivative of the component in the direction of \(\mathbf{k}\), which is given by the product of \(-2t^{-1}\) and \(-3t\). We use the product rule and the chain rule for the reciprocal function: $$(t^{-1})' = -t^{-2}$$ So for our case, we have $$\frac{d}{dt}(-2t^{-1}(-3t)) = 6(-2t^{-2}(-3t)) + (-2t^{-1})(-3)$$
04

Combine the derivatives

Now we combine the derivatives of each component to form the total time derivative of the given function: $$\frac{d}{dt}\left(\left(3t^2 \mathbf{i}+\sqrt{t} \mathbf{j}-2t^{-1} \mathbf{k}\right) \cdot(\cos t \mathbf{i}+\sin 2t \mathbf{j}-3t \mathbf{k})\right) =$$ $$[ (6t\cos t - 3t^2\sin t) \mathbf{i} + (\frac{1}{2\sqrt{t}}(\sin 2t) + \sqrt{t}(2\cos 2t)) \mathbf{j} + (6(-2t^{-2}(-3t)) + (-2t^{-1})(-3)) \mathbf{k} ]$$

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Most popular questions from this chapter

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Geometric-arithmetic mean Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

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