Find parametric equations for the tangent line at the point

In summary, the task was to find parametric equations for the tangent line at a given point on a curve. The first step was to take the derivative of the curve, which resulted in r'(t)=( -sint, cost, 1). To find the correct t value, the equations were compared with the given point, leading to t= 4pi/6. The final tangent line equation was found by using the tangent vector and the point on the curve, giving the equation f(s) = s.r'+p.
  • #1
undrcvrbro
132
0

Homework Statement


Find parametric equations for the tangent line at the point

[tex](\cos(\frac{4 \pi}{6}) ,\sin(\frac{4 \pi}{6}) ,\frac{4 \pi}{6}) )[/tex]

on the curve


[tex]x=\cos t,\ y=\sin t, \ z=t [/tex]

Homework Equations





The Attempt at a Solution


Took the derivative
r'(t)=( -sint, cost, 1)

Here's where I start to fall apart...I'm not sure what to do...but..is this close...

cos(4pi/6)= -1/2

r'(-1/2) = (-sin(-1/2), cos(-1/2), 1)

so,

x= (-1/2) + (-sin(-1/2))t
y= (sin(4pi/6)) + (cos(-1/2)t
z= 4pi/6 + t

So where did I go wrong? I hope I'm not that far off. If anything needs clarified, please ask.
 
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  • #2
undrcvrbro said:

Homework Statement


Find parametric equations for the tangent line at the point

[tex](\cos(\frac{4 \pi}{6}) ,\sin(\frac{4 \pi}{6}) ,\frac{4 \pi}{6}) )[/tex]

on the curve


[tex]x=\cos t,\ y=\sin t, \ z=t [/tex]

Homework Equations





The Attempt at a Solution


Took the derivative
r'(t)=( -sint, cost, 1)

Here's where I start to fall apart...I'm not sure what to do...but..is this close...

cos(4pi/6)= -1/2

r'(-1/2) = (-sin(-1/2), cos(-1/2), 1)
based on what you have said why choose the point t = -1/2? i would think t = 4.pi/6 is a better choice ;) and what the question asks for

undrcvrbro said:
so,

x= (-1/2) + (-sin(-1/2))t
y= (sin(4pi/6)) + (cos(-1/2)t
z= 4pi/6 + t

So where did I go wrong? I hope I'm not that far off. If anything needs clarified, please ask.

then once you have the tangent vector r', and teh point of the curve say p, the equation o teh tangent line will be

f(s) = s.r'+p
 
  • #3
Thank you! I have the right answer, but I don't understand why you chose t= 4pi/6..can you explain?
 
  • #4
compare you original parametric equations with teh point you are asked to evaluate it at, this gives you the correct t value
 
  • #5
lanedance said:
compare you original parametric equations with teh point you are asked to evaluate it at, this gives you the correct t value

Ohhh. Yikes, I can't believe I didn't notice that! Thanks again!
 

FAQ: Find parametric equations for the tangent line at the point

What is the definition of a tangent line?

A tangent line is a line that touches a curve at a single point, without crossing or intersecting the curve at any other point. It represents the instantaneous rate of change of the curve at that point.

Why do we need to find parametric equations for the tangent line?

Parametric equations allow us to describe the motion of a point on a curve in terms of a parameter, such as time. Finding parametric equations for the tangent line allows us to determine the direction and rate of change of the curve at a specific point, which can be useful in various applications such as physics and engineering.

How do we find parametric equations for the tangent line at a given point?

To find parametric equations for the tangent line at a point, we first need to find the slope of the tangent line at that point. This can be done by taking the derivative of the original curve and evaluating it at the given point. Then, we use the point-slope formula to find the equation of the tangent line in the form of y = mx + b, where m is the slope and b is the y-intercept. Finally, we can convert this equation into parametric form by expressing x and y in terms of a parameter, usually t.

What information do we need to find parametric equations for the tangent line at a point?

To find parametric equations for the tangent line at a point, we need to know the coordinates of the point and the equation of the curve. We also need to have a good understanding of derivatives and the point-slope formula.

What is the significance of finding parametric equations for the tangent line at a point?

Finding parametric equations for the tangent line at a point allows us to understand the behavior of a curve at that point, and how it changes over time. This information can be used in various fields such as physics, engineering, and economics to make predictions and solve real-world problems.

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