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Chapter 11: Problem 7

Differentiate the following functions. $$\mathbf{r}(t)=\left\langle\cos t, t^{2}, \sin t\right\rangle$$

Short Answer

Expert verified
Answer: The derivative of the given vector function with respect to \(t\) is \(\frac{d\mathbf{r}}{dt}=\left\langle -\sin t, 2t, \cos t\right\rangle\).

Step by step solution

01

Identify the components of the vector function

We are given a vector function \(\mathbf{r}(t)=\left\langle\cos t, t^{2}, \sin t\right\rangle\). The components of this vector function are: 1. \(x(t) = \cos t\) 2. \(y(t) = t^2\) 3. \(z(t) = \sin t\)
02

Differentiate each component

Next, we will differentiate each component of the vector function with respect to \(t\): 1. \(\frac{dx}{dt} = \frac{d}{dt} (\cos t) = -\sin t\) 2. \(\frac{dy}{dt} = \frac{d}{dt} (t^2) = 2t\) 3. \(\frac{dz}{dt} = \frac{d}{dt} (\sin t) = \cos t\)
03

Write the derivative as a vector

Now, we will write the derivatives we obtained in Step 2 in the form of a vector. This will be the derivative of the given vector function: $$\frac{d\mathbf{r}}{dt} = \left\langle\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\right\rangle = \left\langle -\sin t, 2t, \cos t \right\rangle$$ So, the derivative of the given vector function is: $$\frac{d\mathbf{r}}{dt}=\left\langle -\sin t, 2t, \cos t\right\rangle$$

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Most popular questions from this chapter

Distance between a point and a line in the plane Use projections to find a general formula for the (least) distance between the point \(\left.P\left(x_{0}, y_{0}\right) \text { and the line } a x+b y=c . \text { (See Exercises } 62-65 .\right)\).

Show that two nonzero vectors \(\mathbf{u}=\left\langle u_{1}, u_{2}\right\rangle\) and \(\mathbf{v}=\left\langle v_{1}, v_{2}\right\rangle\) are perpendicular to each other if \(u_{1} v_{1}+u_{2} v_{2}=0\)

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