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Chapter 12: Problem 13

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

Short Answer

Expert verified
Question: Find the derivative of the vector function \(\mathbf{r}(t)=\left\langle t e^{-t}, t \ln t, t \cos t\right\rangle\). Answer: The derivative of the vector function is \(\mathbf{r}'(t)=\left\langle e^{-t}(1-t), \ln t + 1, \cos t - t\sin t\right\rangle\).

Step by step solution

01

Identify the function components

The given function has three components: \(t e^{-t}\), \(t \ln t\), and \(t \cos t\). We'll differentiate each of those components with respect to \(t\).
02

Differentiate the first component

Let's differentiate $$f_1(t)=t e^{-t}.$$Notice that this function is the product of \(t\) and \(e^{-t}\). To differentiate this, we can apply the product rule: $$(uv)' = u'v + uv'$$. Differentiating each part, we have: $$u = t \Rightarrow u' = 1$$ $$v = e^{-t} \Rightarrow v' = -e^{-t}$$ So, \(f_1'(t) = u'v + uv' = (1)(e^{-t}) + (t)(-e^{-t}) = e^{-t}(1-t)\).
03

Differentiate the second component

Next, we differentiate $$f_2(t) = t\ln t$$, which is the product of \(t\) and \(\ln t\). Using the product rule again, we differentiate each part: $$u = t \Rightarrow u' = 1$$ $$v = \ln t \Rightarrow v' = \frac{1}{t}$$ Now apply the product rule: $$f_2'(t) = u'v + uv' = (1)(\ln t) + (t)\left(\frac{1}{t}\right) = \ln t + 1$$.
04

Differentiate the third component

Lastly, let's differentiate $$f_3(t)=t\cos t.$$ This function is also a product, of \(t\) and \(\cos t\). We differentiate each part: $$u = t \Rightarrow u' = 1$$ $$v = \cos t \Rightarrow v' = -\sin t$$ And apply the product rule: $$f_3'(t) = u'v + uv' = (1)(\cos t) + (t)(-\sin t) = \cos t - t\sin t$$.
05

Combine the differentiated components

Now that we have the derivatives of all three components, we combine them to form the derivative of the vector function \(\mathbf{r}'(t)\): $$\mathbf{r}'(t)=\left\langle f_1'(t), f_2'(t), f_3'(t)\right\rangle = \left\langle e^{-t}(1-t), \ln t + 1, \cos t - t\sin t\right\rangle$$ Thus, the derivative of the given function is: $$\mathbf{r}'(t)=\left\langle e^{-t}(1-t), \ln t + 1, \cos t - t\sin t\right\rangle.$$

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