Find parametric equations for the tangent line to the curve

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Homework Help Overview

The discussion revolves around finding parametric equations for the tangent line to a curve defined by a vector function. Participants are analyzing derivatives and attempting to determine the correct values of parameters at a specific point.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivatives of the vector components and their evaluations at the point (5, 2, 0). Questions arise regarding the correct value of the parameter t that corresponds to this point on the curve.

Discussion Status

Some participants have offered insights into the need to solve for t using the original equations rather than the derivatives. There is an acknowledgment of a misunderstanding regarding the evaluation of the derivatives at the specified point.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is a specific focus on ensuring that the equations accurately represent the curve at the given point.

Atlos
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Homework Statement



symimage.gif

Homework Equations



r = <x,y,z>
r' = <x',y',z'>

The Attempt at a Solution



I started by finding the derivatives of each part of the vector and got:

x= 2/sqrt(t) y= 3t^2+1 z= 3t^2-1

Then I plugged the point (5,2,0) into that and got (2/sqrt(5), 13, -1). This should be multiplied by t. Then by plugging in (5,2,0) back into the original line, you get (1+4sqrt(5), 10, 0). I thought that the parametric equation for this would be (1+4sqrt(5), 10, 0) + (2/sqrt(5), 13, -1)*t, but it's not working. The solution y(t) = 2+4t is already given, so I'm obviously already wrong. Any idea what I did wrong?
 
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Atlos said:

Homework Statement



symimage.gif



Homework Equations



r = <x,y,z>
r' = <x',y',z'>

The Attempt at a Solution



I started by finding the derivatives of each part of the vector and got:

x= 2/sqrt(t) y= 3t^2+1 z= 3t^2-1
These would be x' = 2/sqrt(t), y'= 3t^2+1, z'= 3t^2-1

Atlos said:
Then I plugged the point (5,2,0) into that and got (2/sqrt(5), 13, -1).
Each of these derivatives is a function of t. At the point (5, 2, 0), what's the value of t?
Atlos said:
This should be multiplied by t. Then by plugging in (5,2,0) back into the original line, you get (1+4sqrt(5), 10, 0). I thought that the parametric equation for this would be (1+4sqrt(5), 10, 0) + (2/sqrt(5), 13, -1)*t, but it's not working. The solution y(t) = 2+4t is already given, so I'm obviously already wrong. Any idea what I did wrong?
 
Mark44 said:
These would be x' = 2/sqrt(t), y'= 3t^2+1, z'= 3t^2-1

Each of these derivatives is a function of t. At the point (5, 2, 0), what's the value of t?

Woops, I forgot the prime for those. What do you mean what would the value of t be? Like solve for 2/sqrt(t) = 5, 3t^2+1 = 2, and 3t^2-1 = 0 to find a t vector? Otherwise I don't see a value of t that would make the equations go through the point (5,2,0).

edit: oh nevermind t would be 1 to get to the point (5,2,0). So should I plug 1 into the derivative for each?
 
Solve for t in the equations for x, y, and z, at the point (5, 2, 0), NOT the equations for x', y', and z'. (5, 2, 0) is a point on the curve described by the parametric equations.
 
Ah, got it! I understand it now, I got x(t)=5+2t and z(t)=2t which were both right.
 

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