Parametric Equations - Have solution, need clarification.

Since both 1+ 2\sqrt{t}= 3 and t^4+ t= 2 are satisfied, (1+ 2\sqrt{t}, t^4+ t, t^4- t)= (3, 2, 0) when t= 1. That is, the point (3, 2, 0) is on the curve when t= 1. Now, you are asked to find the tangent line at that point. You found the tangent vector at t= 1 to be <1/3, 4, 0> (since dr/dt= <1+ 1/\sqrt{t}, 4t^3+
  • #1
tangibleLime
71
0

Homework Statement


Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

x = 2cos(t)
y = sin(t)
z = t

At the point (0,1,pi/2)

Homework Equations


The Attempt at a Solution



r(t) = <2cos(t),sin(t),t>
r'(t) = <-2sin(t),cos(t),1>

The parameter value corresponding to the point (0,1,pi/2) is t=pi/2, so the tangent vector there is r'(pi/2) = <-2,0,1>. The tangent line through (0,1,pi/2) parallel to the vector <-2,0,1>, so its parametric equations are:

x=-2t
y=1
z = (pi/2)+t

I know this is correct, but the part in bold is what I do not understand. Why does the value of t=pi/2 correspond to the point (0,1,pi/2)?
 
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  • #2
tangibleLime said:

Homework Statement


Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

x = 2cos(t)
y = sin(t)
z = t

At the point (0,1,pi/2)

Homework Equations





The Attempt at a Solution



r(t) = <2cos(t),sin(t),t>
r'(t) = <-2sin(t),cos(t),1>

The parameter value corresponding to the point (0,1,pi/2) is t=pi/2, so the tangent vector there is r'(pi/2) = <-2,0,1>. The tangent line through (0,1,pi/2) parallel to the vector <-2,0,1>, so its parametric equations are:

x=-2t
y=1
z = (pi/2)+t

I know this is correct, but the part in bold is what I do not understand. Why does the value of t=pi/2 correspond to the point (0,1,pi/2)?

What happens if you calculate r(pi/2) in r(t) = <2cos(t),sin(t),t>? Would any other value of t work in the 3rd component?
 
  • #3
Ooh okay, makes sense. Need to just find the component that makes r(t) equal the given point. Thanks!
 
  • #4
Actually no, that doesn't work for my problem.

(Now I'm doing the actual question I have to solve instead of the book example)

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

x = 1+2sqrt(t)
y = t^4 + t
z = t^4 - t

(3,2,0)

I have r(t) = <1+2sqrt(t),t^4 + t,t^4 - t>, but no point (3, 2 or 0) will make this vector (3, 2, 0). What am I doing wrong?
 
  • #5
Have you tried t=1?
 
  • #6
Oh, so I can pick ANY number that makes r(t) equal to the given point?
 
  • #7
well, yeah. Although if you have multiple t that make r(t) fit the point, you've got a pretty cool function.
 
  • #8
In parametric equations, x= f(t), y= g(t), z= h(t), the parameter, t, is not necessarily equal to any of x, y, or z. It is a completely different parameter.


Sometimes, back in two dimensions, if we have, say y= f(x), we use x itself as parameter. That may have confused you.

Here, you want [itex](1+2\sqrt(t),t^4 + t,t^4 - t)= (3, 2, 0)[/itex] so we must have [itex]1+ 2\sqrt{t}= 3[/itex], [itex]t^4+ t= 2[/itex], and [itex]t^4- t= 0[/itex] all for the same t. We could, for example, solve [itex]t^4- t= t(t^3- 1)= t(t- 1)(t^2+ t+ 1)= 0[/itex] to see that t must be 0 or 1. t= 0 does not satisfy either [itex]1+ 2\sqrt{t}= 3[/itex] nor [itex]t^4+ t= 2[/itex] but t= 1 does.
 

1. What are parametric equations?

Parametric equations are a type of mathematical representation of a curve or surface in terms of one or more parameters. They allow for a more flexible and intuitive way of describing curves and surfaces compared to traditional Cartesian equations.

2. How are parametric equations different from Cartesian equations?

Parametric equations use parameters to describe the relationship between two variables, while Cartesian equations directly relate the variables to each other. Parametric equations can also describe curves and surfaces that are difficult to express using Cartesian equations.

3. How do you graph parametric equations?

To graph parametric equations, you can plot points by substituting different values for the parameters and then connecting the points with a smooth curve. Alternatively, you can use a graphing calculator or software to plot the equations.

4. What is the importance of parametric equations in science?

Parametric equations are used in various fields of science, such as physics, engineering, and computer graphics, to describe and model complex curves and surfaces. They allow for a more precise and efficient way of representing these objects compared to traditional equations.

5. Can parametric equations have multiple solutions?

Yes, parametric equations can have multiple solutions, just like any other mathematical equation. The number of solutions depends on the number of parameters and the constraints placed on them. In some cases, there may be an infinite number of solutions.

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