rzfansubs87

Answered

2022-07-06

How do you find parametric equations for the tangent line to the curve with the given parametric equations at the point $(1,4,4)$ .

$x=\mathrm{cos}\left(t\right),y=4{e}^{8}t,z=4{e}^{-8}t$

$x=\mathrm{cos}\left(t\right),y=4{e}^{8}t,z=4{e}^{-8}t$

Answer & Explanation

enfeinadag0

Expert

2022-07-07Added 16 answers

Step 1

We have parametric equations as functions of t for x, y and z so we can form the derivatives wrt t for each variable:

$x=\mathrm{cos}t\Rightarrow \frac{dx}{dt}=-\mathrm{sin}t$

$y=4{e}^{8t}\Rightarrow \frac{dy}{dt}=32{e}^{8t}$

$z=4{e}^{-8t}\Rightarrow \frac{dz}{dt}=-32{e}^{8t}$

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

$\nabla f\left(t\right)=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k}$

$=-\mathrm{sin}t\hat{i}+32{e}^{8t}\hat{j}-32{e}^{-8t}\hat{k}$

So for the particular point $(1,4,4)$ we need to establish the corresponding value of t, which by inspection is $t=0$ .

So for this particular point, the normal vector to the surface is given by:

$\nabla f\left(0\right)=0\hat{i}+32\hat{j}-32\hat{k}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=32\hat{j}-32\hat{k}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=32(\hat{j}-\hat{k})$So the tangent line is a line with direction $\hat{j}-\hat{k}$ that passes through the point $(1,4,4)$ , which therefore has the vector equation:

$\overrightarrow{r}=\left(\begin{array}{c}1\\ 4\\ 4\end{array}\right)+\lambda \left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)$

Which we can therefore parametrise (using $\lambda$ ) as follows

$x=1$

$y=4+\lambda$

$z=4-\lambda$

We have parametric equations as functions of t for x, y and z so we can form the derivatives wrt t for each variable:

$x=\mathrm{cos}t\Rightarrow \frac{dx}{dt}=-\mathrm{sin}t$

$y=4{e}^{8t}\Rightarrow \frac{dy}{dt}=32{e}^{8t}$

$z=4{e}^{-8t}\Rightarrow \frac{dz}{dt}=-32{e}^{8t}$

In order to find the normal at any particular point in vector space we use the Del, or gradient operator:

$\nabla f\left(t\right)=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}+\frac{dz}{dt}\hat{k}$

$=-\mathrm{sin}t\hat{i}+32{e}^{8t}\hat{j}-32{e}^{-8t}\hat{k}$

So for the particular point $(1,4,4)$ we need to establish the corresponding value of t, which by inspection is $t=0$ .

So for this particular point, the normal vector to the surface is given by:

$\nabla f\left(0\right)=0\hat{i}+32\hat{j}-32\hat{k}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=32\hat{j}-32\hat{k}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=32(\hat{j}-\hat{k})$So the tangent line is a line with direction $\hat{j}-\hat{k}$ that passes through the point $(1,4,4)$ , which therefore has the vector equation:

$\overrightarrow{r}=\left(\begin{array}{c}1\\ 4\\ 4\end{array}\right)+\lambda \left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)$

Which we can therefore parametrise (using $\lambda$ ) as follows

$x=1$

$y=4+\lambda$

$z=4-\lambda$

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