Finding Tangent and Intersection for Parametric Curve

In summary, the conversation discusses finding the equation of a tangent to a parametric curve and finding the value of a parameter for the second point of intersection. The key steps involve substituting values for t and using the derivative to find the equation of the tangent. Additionally, the condition for a tangent to intersect a parametric curve is discussed.
  • #1
Seydlitz
263
4

Homework Statement


A curve is defined by the parametric equations:

x = 2t^3
y = 2t^2
t =/ 0

1)Prove that the equation of the tangent at the point with parameter t is 2x - 3ty + 2t^3 = 0. Proven, and I've no problem with this part.

2.)The tangent at the point t = 2 meets the curve again at the point where t = u. Find the value of u.

The Attempt at a Solution



I substituted the value of t = 2 to the first equation in problem 1.
It gives me 2x - 6y + 16 = 0
y = (1/6)(2x + 16)

At this point I'm stuck because ordinarily I'll just set the tangent equation equal to the curve, expressed in terms of x and then solve for x. But now with parametric equation I do not know how to proceed next or how to see the problem.

What is the condition required for a tangent to be said intersecting a parametric curve?

Thank You
 
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  • #2
Update: After a long moment of gedankenexperiment, :D

I've decided to substitute the initial curve parameter to the equation of the tangent with t = 2. That is 2x - 6y + 16 = 0. x = 2t^3, y = 2t^2. I already know one of the root and proceed to factorize the equation. Which gives me the solution, t = 2 and t = -1. So u = -1.
 
  • #3
At t=u, find the parametric equations for x and y. Then replace in your equation of tangent for t=2.

But what you've done is correct as well. The solution for "u" might not have been so obvious if the question was not simply t=u, for example, t=(2u-4).
 
  • #4
Is my solution acceptable? (2nd post)
 
  • #5
Remember. [itex]\displaystyle\dfrac{\mathrm{d}y}{\mathrm{d}x}= \dfrac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{ \mathrm{d}x}{\mathrm{d}t}}[/itex]. It turns out that the d's are actual terms, but most analysis courses don't teach that and tell us to realize this isn't "proper" math because understanding them is enormously difficult. So always keep in mind that you _can_ treat them like fractions. Just don't tell your teacher. :)
 
  • #6
Seydlitz said:
What is the condition required for a tangent to be said intersecting a parametric curve?

If the 2nd point of intersection is (A,B), then if you put x=A into the equation of tangent, making y the subject of formula, you should get the answer B (that is, the y-coordinate of the 2nd point).
 
  • #7
Whovian said:
Remember. [itex]\displaystyle\dfrac{\mathrm{d}y}{\mathrm{d}x}= \dfrac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{ \mathrm{d}x}{\mathrm{d}t}}[/itex]. It turns out that the d's are actual terms, but most analysis courses don't teach that and tell us to realize this isn't "proper" math because understanding them is enormously difficult. So always keep in mind that you _can_ treat them like fractions. Just don't tell your teacher. :)

Ahh that's the case! I've been doing it a bit unconsciously with parametric problem then suddenly I know there's deep secret underneath it. Pretty neat.

sharks said:
If the 2nd point of intersection is (A,B), then if you put x=A into the equation of tangent, making y the subject of formula, you should get the answer B (that is, the y-coordinate of the 2nd point).

Alright, I've checked my solution accordingly. Thank You!
 

FAQ: Finding Tangent and Intersection for Parametric Curve

1. What is a parametric curve?

A parametric curve is a type of mathematical curve that is described by a set of equations called parametric equations. These equations use one or more parameters to define the coordinates of points on the curve.

2. What is the purpose of finding tangent and intersection for a parametric curve?

Finding tangent and intersection points on a parametric curve allows us to determine the slope and the points of intersection with other curves or lines. This information is important in many fields, including physics, engineering, and computer graphics.

3. How do you find the tangent of a parametric curve?

To find the tangent of a parametric curve at a specific point, we use the derivative of the parametric equations. We can find the derivative by using the chain rule or by eliminating the parameter and finding the derivative with respect to the independent variable.

4. What is the process for finding intersection points of two parametric curves?

To find the intersection points of two parametric curves, we first equate the equations for each curve and solve for the parameters. Then, we substitute these values back into the original equations to find the corresponding coordinates of the intersection points.

5. Are there any other methods for finding tangent and intersection points of parametric curves?

Yes, there are other methods for finding tangent and intersection points of parametric curves, such as using polar coordinates or converting the parametric equations into Cartesian equations. However, the most common and efficient method is using derivatives and solving equations algebraically.

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