Question

The graph of the vector-valued function \(\mathbf{r}(t)\) and a tangent vector to the graph at \(t=t_{0}\) are given. (a) Find a set of parametric equations for the tangent line to the graph at \(t=t_{0}\) (b) Use the equations for the tangent line to approximate \(\mathbf{r}\left(t_{0}+\mathbf{0 . 1}\right)\) $$ \mathbf{r}(t)=\left\langle t,-t^{2}, \frac{1}{4} t^{3}\right\rangle, \quad t_{0}=1 $$

Short answer

The set of parametric equations for the tangent line to the graph at \(t=1\) is \(x(t)=1+(1)(t-1)\), \(y(t)=-1+(-2)(t-1)\), \(z(t)=\frac{1}{4}+\left(\frac{3}{4}\right)(t-1)\). The approximation \(\mathbf{r}(1.1)\) using these equations is approximately \(\left\langle 1.1,-1.2,\frac{5}{8}\right\rangle\).

Step-by-step solution

Compute the derivative of \(\mathbf{r}(t)\)

The derivative of \(\mathbf{r}(t)=\left\langle t,-t^{2}, \frac{1}{4} t^{3}\right\rangle\) can be calculated component-wise, leading to: \[\mathbf{r}'(t)=\left\langle 1,-2t, \frac{3}{4} t^{2}\right\rangle.\]

Evaluate the derivative at \(t=t_{0}\)

We are given that \(t_{0}=1\). Therefore, the value of \(\mathbf{r}'(t)\) at \(t_{0}=1\) gives the direction of the tangent vector: \[\mathbf{r}'(1)=\left\langle 1,-2(1), \frac{3}{4} (1)^{2}\right\rangle=\left\langle 1,-2,\frac{3}{4}\right\rangle.\]

Compute the value of \(\mathbf{r}(t)\) at \(t=t_{0}\)

The value of \(\mathbf{r}(t)\) at \(t_{0}=1\) gives the point where the tangent hits the graph: \[\mathbf{r}(1)=\left\langle 1,-(1)^2, \frac{1}{4} (1)^3\right\rangle=\left\langle 1,-1,\frac{1}{4}\right\rangle.\]

Write the parametric equations for the tangent line

With the point \(\mathbf{r}(1)\) and the direction vector \(\mathbf{r}'(1)\) in hand, the tangent line \(\mathbf{l}(t)\) to the graph at \(t=1\) can be written as: \[\mathbf{l}(t)=\mathbf{r}(1)+\mathbf{r}'(1)(t-1).\] Thus the set of parametric equations is: \[\begin{aligned} x(t)&= 1+(1)(t-1)\ y(t)&= -1+(-2)(t-1)\ z(t)&= \frac{1}{4}+\left(\frac{3}{4}\right)(t-1) \end{aligned}\]

Compute the value of \(\mathbf{l}(t)\) at \(t=t_{0}+0.1\)

We can approximate \(\mathbf{r}\left(t_{0}+\mathbf{0 . 1}\right)\) using the tangent line \(\mathbf{l}(t)\) evaluated at \(t=t_{0}+0.1\). Therefore, replace \(t\) by \(1.1\) in each of the parametric equations: \[\begin{aligned} x(1.1)&= 1+(1)(1.1-1)\ y(1.1)&= -1+(-2)(1.1-1)\ z(1.1)&= \frac{1}{4}+\left(\frac{3}{4}\right)(1.1-1) \end{aligned}.\] This gives us the approximation \(\mathbf{r}(1.1)\approx\mathbf{l}(1.1).\]

Simplify the results

The simplified results for Step 5 are: \[\begin{aligned} x(1.1)&= 1.1\ y(1.1)&= -1.2\ z(1.1)&= \frac{5}{8} \end{aligned}\] Therefore, \(\mathbf{r}(1.1)\approx\mathbf{l}(1.1)=\left\langle 1.1,-1.2,\frac{5}{8}\right\rangle.\]