Find parametric equations for the tangent line to the curve

In summary, the student attempted to solve for t but forgot to include the prime for t in the equation, and then got confused when t turned out to be 1.
  • #1
Atlos
11
0

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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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
Ah, got it! I understand it now, I got x(t)=5+2t and z(t)=2t which were both right.
 

What are parametric equations for tangent line to a curve?

Parametric equations for the tangent line to a curve are equations that describe the x and y coordinates of points on the line as a function of a third variable, usually denoted as t. These equations can be used to find the slope of the tangent line at any point on the curve.

How do you find the slope of the tangent line using parametric equations?

To find the slope of the tangent line at a specific point on the curve, you can use the derivative of the parametric equations. The derivative will give you the rate of change of y with respect to x, which is the slope of the tangent line.

Can parametric equations be used for any type of curve?

Yes, parametric equations can be used for any type of curve, including circles, ellipses, and more complex curves. This is because the equations describe the coordinates of points on the curve as a function of a third variable, allowing for a wide range of curves to be represented.

How do you find the parametric equations for a specific curve?

The process for finding parametric equations for a curve can vary depending on the type of curve. In general, you can start by expressing the curve in the form of x = f(t) and y = g(t), where f and g are functions of t. From there, you can solve for t and substitute it into the equations to get the parametric equations for the curve.

Are parametric equations the only way to find the tangent line to a curve?

No, parametric equations are not the only way to find the tangent line to a curve. Other methods include using the slope formula or the derivative of the function representing the curve. However, parametric equations can be useful for more complex curves that cannot be easily represented by a single function.

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