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Chapter 12: Problem 10

Differentiate the following functions. $$\mathbf{r}(t)=\langle 4,3 \cos 2 t, 2 \sin 3 t\rangle$$

Short Answer

Expert verified
Question: Find the derivative of the given vector function with respect to t: $$\mathbf{r}(t) = \langle 4, 3 \cos 2 t, 2 \sin 3 t \rangle .$$ Answer: $$\frac{d\mathbf{r}}{dt} = \langle 0, -6 \sin 2t, 6 \cos 3t \rangle .$$

Step by step solution

01

Identify Components of the Vector Function

The given vector function is $$\mathbf{r}(t) = \langle 4, 3 \cos 2 t, 2 \sin 3 t \rangle .$$ The components are: $$ x(t) = 4 ,$$ $$ y(t) = 3 \cos 2t ,$$ $$ z(t) = 2 \sin 3t .$$
02

Differentiate the x(t) component

The x(t) component is a constant, so its derivative is: $$\frac{dx}{dt} = 0.$$
03

Differentiate the y(t) component

The y(t) component is a cosine function multiplied by a constant. We will use the chain rule for differentiation: $$\frac{dy}{dt} = \frac{d(3 \cos 2t)}{dt} =3 \frac{d(\cos 2t)}{dt}.$$ Now, differentiate the cosine function using the chain rule: $$= 3 (-\sin 2t) \frac{d(2t)}{dt}$$ $$= -6 \sin 2t.$$ So, we have $$\frac{dy}{dt} =-6 \sin 2t.$$
04

Differentiate the z(t) component

The z(t) component is a sine function multiplied by a constant. Again, we use the chain rule for differentiation: $$\frac{dz}{dt} = \frac{d(2 \sin 3t)}{dt} = 2 \frac{d(\sin 3t)}{dt}.$$ Now differentiate the sine function using the chain rule: $$= 2 (\cos 3t) \frac{d(3t)}{dt}$$ $$= 6 \cos 3t.$$ So, we have $$\frac{dz}{dt} = 6 \cos 3t.$$
05

Combine the Results

In the final step, combine the results of the differentiated components into a single vector. The derivative of the vector function is: $$\frac{d\mathbf{r}}{dt} = \langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \rangle = \langle 0, -6 \sin 2t, 6 \cos 3t \rangle .$$

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