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Chapter 11: Problem 16

Find a tangent vector at the given value of \(t\) for the following parameterized curves. $$\mathbf{r}(t)=\left\langle e^{t}, e^{3 t}, e^{5 t}\right\rangle, t=0$$

Short Answer

Expert verified
Answer: The tangent vector to the curve at \(t=0\) is \(\langle 1, 3, 5 \rangle\).

Step by step solution

01

Calculate the derivative of the given vector function with respect to \(t\)

The derivative of a vector function \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle\) with respect to \(t\) is found by taking the derivatives of each component with respect to \(t\). We have: $$\frac{d\mathbf{r}}{dt} = \frac{d\langle e^t, e^{3t}, e^{5t} \rangle}{dt} = \langle \frac{de^t}{dt}, \frac{de^{3t}}{dt}, \frac{de^{5t}}{dt} \rangle$$ Using the chain rule for differentiation, we get: $$\frac{d\mathbf{r}}{dt} = \langle e^t , 3 e^{3t} , 5 e^{5t} \rangle$$
02

Substitute the given value \(t=0\) into the expression for the derivative

We will now substitute \(t=0\) into the expression for the derivative we found in step 1: $$\left.\frac{d\mathbf{r}}{dt}\right|_{t=0} = \langle e^0 , 3 e^{3(0)} , 5 e^{5(0)} \rangle$$ Since \(e^0=1\), we have: $$\left.\frac{d\mathbf{r}}{dt}\right|_{t=0} = \langle 1, 3, 5 \rangle$$
03

Interpret the result

The tangent vector we found, \(\langle 1, 3, 5 \rangle\), represents the direction of the tangent to the curve \(\mathbf{r}(t)\) at the point where \(t=0\). This means that if we move along the curve in the direction specified by the tangent vector starting from the point \(t=0\), we will be moving "along" the curve while staying as close to the curve as possible.

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Most popular questions from this chapter

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and \(f\) are real numbers. It can be shown that this curve lies in a plane. Graph the following curve and describe it. $$\begin{aligned}\mathbf{r}(t)=&\left(\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{i}+\left(-\frac{1}{\sqrt{2}} \cos t+\frac{1}{\sqrt{3}} \sin t\right) \mathbf{j} \\\&+\left(\frac{1}{\sqrt{3}} \sin t\right) \mathbf{k} \end{aligned}$$

Let \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\). a. Assume that \(\lim \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3}\right\rangle,\) which means that \(\lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0 .\) Prove that \(\lim _{t \rightarrow a} f(t)=L_{1}, \quad \lim _{t \rightarrow a} g(t)=L_{2}, \quad\) and \(\quad \lim _{t \rightarrow a} h(t)=L_{3}\). b. Assume that \(\lim _{t \rightarrow a} f(t)=L_{1}, \lim _{t \rightarrow a} g(t)=L_{2},\) and \(\lim _{t \rightarrow a} h(t)=L_{3} .\) Prove that \(\lim _{t \rightarrow a} \mathbf{r}(t)=\mathbf{L}=\left\langle L_{1}, L_{2}, L_{3}\right\rangle\) which means that \(\lim _{t \rightarrow a}|\mathbf{r}(t)-\mathbf{L}|=0\).

Use vectors to show that the midpoint of the line segment joining \(P\left(x_{1}, y_{1}\right)\) and \(Q\left(x_{2}, y_{2}\right)\) is the point \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) (Hint: Let \(O\) be the origin and let \(M\) be the midpoint of \(P Q\). Draw a picture and show that $$\left.\overrightarrow{O M}=\overrightarrow{O P}+\frac{1}{2} \overrightarrow{P Q}=\overrightarrow{O P}+\frac{1}{2}(\overrightarrow{O Q}-\overrightarrow{O P}) \cdot\right)$$

Prove the following vector properties using components. Then make a sketch to illustrate the property geometrically. Suppose \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in the \(x y\) -plane and a and \(c\) are scalars. $$\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$$

Compute the following derivatives. $$\frac{d}{d t}\left(t^{2}(\mathbf{i}+2 \mathbf{j}-2 t \mathbf{k}) \cdot\left(e^{t} \mathbf{i}+2 e^{t} \mathbf{j}-3 e^{-t} \mathbf{k}\right)\right)$$
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