rzfansubs87

2022-07-06

How do you find parametric equations for the tangent line to the curve with the given parametric equations at the point $\left(1,4,4\right)$ .
$x=\mathrm{cos}\left(t\right),y=4{e}^{8}t,z=4{e}^{-8}t$

Expert

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:

$z=4{e}^{-8t}⇒\frac{dz}{dt}=-32{e}^{8t}$
In order to find the normal at any particular point in vector space we use the Del, or gradient operator:
$\nabla f\left(t\right)=\frac{dx}{dt}\stackrel{^}{i}+\frac{dy}{dt}\stackrel{^}{j}+\frac{dz}{dt}\stackrel{^}{k}$

So for the particular point $\left(1,4,4\right)$ 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:
$\nabla f\left(0\right)=0\stackrel{^}{i}+32\stackrel{^}{j}-32\stackrel{^}{k}$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=32\stackrel{^}{j}-32\stackrel{^}{k}$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=32\left(\stackrel{^}{j}-\stackrel{^}{k}\right)$So the tangent line is a line with direction $\stackrel{^}{j}-\stackrel{^}{k}$ that passes through the point $\left(1,4,4\right)$ , which therefore has the vector equation:
$\stackrel{\to }{r}=\left(\begin{array}{c}1\\ 4\\ 4\end{array}\right)+\lambda \left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)$
Which we can therefore parametrise (using $\lambda$ ) as follows
$x=1$
$y=4+\lambda$
$z=4-\lambda$

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