Equation of Tangent Line to Curve at Point

  • #1

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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619

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.
If I put t=0 into -4e^(-4t) I get -4*e^0=(-4).
 
  • #3
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,555
766

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]
Right. So ##\vec r'(0) = \langle 1, -4, 5\rangle##. That gives you the direction vector of that tangent line for ##t=0##.

[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.
After you correct it, you still won't agree because their last component is wrong; it should be ##5t##.

[Edit] Actually, it's your last component that is also wrong.
 
  • #4
Oh.

I was plugging in the components of the vector, not the parameter (so plugging 1 into the y-equation). I don't know why. Thinking and typing one thing, doing a completely different thing, and was totally unaware of it.

-______________________-

Thanks, guys.
 

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