Question

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

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

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 $$\frac{d}{dt}(3t^2\cos t) = (6t\cos t) - (3t^2\sin t)$$

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 $$\frac{d}{dt}(\sqrt{t}\sin 2t) = \frac{1}{2\sqrt{t}}(\sin 2t) + \sqrt{t}(2\cos 2t)$$

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 $$\frac{d}{dt}(-2t^{-1}(-3t)) = 6(-2t^{-2}(-3t)) + (-2t^{-1})(-3)$$

Combine the derivatives

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