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Chapter 10: Problem 11

Find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=a \cos ^{3} t \mathbf{i}+a \sin ^{3} t \mathbf{j}+\mathbf{k} $$

Short Answer

Expert verified
\(\mathbf{r}^{\prime}(t)=-3a\sin(t)\cos^2(t)\mathbf{i} + 3a\sin^2(t)\cos(t)\mathbf{j} + 0\mathbf{k}\)

Step by step solution

01

Differentiate the \(\mathbf{i}\) Component

Firstly, differentiate the component alongside the \(\mathbf{i}\) vector. This involves evaluating the derivative of \(a \cos ^{3} t\) with respect to \(t\). Use the chain rule here, which states that the derivative of a composition of functions is the product of the derivative of the inner function and the derivative of the outer function. The derivative is \(-3a\sin(t)\cos^2(t)\).
02

Differentiate the \(\mathbf{j}\) Component

Now, differentiate the component alongside the \(\mathbf{j}\) vector. Once again, applying the chain rule to \(a \sin ^{3} t\), we will get the derivative as \(3a\sin^2(t)\cos(t)\).
03

Differentiate the \(\mathbf{k}\) Component

Lastly, differentiate the component alongside the \(\mathbf{k}\) vector. As this is a constant term, its derivative is zero.

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Most popular questions from this chapter

What is known about the speed of an object if the angle between the velocity and acceleration vectors is (a) acute and (b) obtuse?

Find the vectors \(\mathrm{T}\) and \(\mathrm{N},\) and the unit binormal vector \(\mathbf{B}=\mathbf{T} \times \mathbf{N},\) for the vector-valued function \(\mathbf{r}(t)\) at the given value of \(t\). $$ \mathbf{r}(t)=2 e^{t} \mathbf{i}+e^{t} \cos t \mathbf{j}+e^{t} \sin t \mathbf{k}, \quad t_{0}=0 $$

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