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## Homework Statement

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

## Homework Equations

[tex]x = t \\

y = e^{-4t} \\

z = 5t - t^5 \\

P = (0, 1, 0)[/tex]

## The Attempt at a Solution

[tex] \vec{r}(t) = < t, e^{-4t}, 5t - t^5 > [/tex]

At the point (0, 1, 0), t = 0.

[tex] \vec{r '}(t) = < 1, -4e^{-4t}, 5 - 5t^4 > [/tex]

[tex] x' = 1 \\

y ' = -4e^{-4t} \\

z ' = 5 - 5t^4[/tex]

These are the x, y, and z equations for the rate of change of the curve in the x, y, and z directions respectively (

*or*, the slopes/direction of the tangent lines at the point when t = some number).

At t = 0, the x, y, and z slopes of the tangent lines are:

[tex] x' = 1 \\

y' = -4e^{-4} \\

z' = 4[/tex]

A point on the single tangent line of the curve (found by adding the vectors x', y', and z') is (0, 1, 0)—since the curve touches that point, too. The direction vector of the single tangent line is < 1, -4(e

^{-4}), 4 >. The parameter is t.

So the vector equation of the tangent line of the curve at the point (0, 1, 0) is:

[tex] (x(t), y(t), z(t)) = (0, 1, 0) + t<1, \frac{-4}{E^{4}}, 4> = < t, 1 -\frac{4t}{E^{4}}, 4t >[/tex]

This, according to Webassign, is incorrect.

Their answer the same, except their y-component is 1 -4t. They... dropped the exponential in the denominator? I can't figure out why that is the correct answer, and why my answer is incorrect.