<meta http-equiv="refresh" content="1; url=/nojavascript/"> Parametric-Inverses ( Read ) | Calculus | CK-12 Foundation
Dismiss
Skip Navigation

Parametric-Inverses

%
Progress
Practice
Progress
%
Practice Now
Parametric Inverses

You have learned that a graph and its inverse are reflections of each other across the line y=x.  You have also learned that in order to find an inverse algebraically, you can switch the x  and  y  variables and solve for  y.  Parametric equations actually make finding inverses easier because both the  x  and  y  variables are based on a third variable  t.  All you need to do to find the inverse of a set of parametric equations and switch the functions for  x  and   y.  

Is the inverse of a function always a function?

Watch This

http://www.youtube.com/watch?v=4Y14XhPD7Os James Sousa: Graphing Parametric Equations in the TI84

Guidance

To find the inverse of a parametric equation you must switch the function of x with the function of y. This will switch all the points from (x,y) to (y,x) and also has the effect of visually reflecting the graph over the line y=x.

Example A

Find and graph the inverse of the parametric function on the domain -2 < t< 2.

x &=2t \\y &=t^2-4

Solution: Switch the  x  and  y  functions and graph. 

x &=t^2-4 \\y &=2t

The original function is shown in blue and the inverse is shown in red.

Example B

Is the point (4, 8) in the following function or its inverse?

x  &=2t^2-2 \\y &=t^2-1

Solution: Try to solve for a matching  t in the original function.

x=2t^2-2

4=2t^2-2

6=2t^2

3=t^2

\pm \sqrt{3}=t

y=t^2-1

8=t^2-1

9=t^2

\pm 3=t

The point does not satisfy the original function.  Check to see if it satisfies the inverse.

x=t^2-1

4=t^2-1

\pm\sqrt{5}=t

y=2t^2-2

8=2t^2-2

10=2t^2

\pm\sqrt{5}=t

The point does satisfy the inverse of the function.

Example C

Parameterize the following function and then graph the function and its inverse.

f(x)=x^2+x-4

Solution: For the original function, the parameterization is:

x &=t \\y &=t^2+t-4

The inverse is:

x &=t^2+t-4 \\y &=t

Concept Problem Revisited

The inverse of a function is not always a function.  In order to see whether the inverse of a function will be a function, you must perform the horizontal line test on the original function.  If the function passes the horizontal line test then the inverse will be a function.  If the function does not pass the horizontal line test then the inverse produces a relation rather than a function.

Vocabulary

Two functions are  inverses  if for every point  (a, b)  on the first function there exists a point  (b, a)  on the second function.

An intersection for two sets of parametric equations happens when the points exist at the same  x, y  and  t.

Guided Practice

1. Find the points of intersection of the function and its inverse from Example C.

2. Does the point (-2, 6) live on the following function or its inverse?

x=t^2-10

y =\frac{t}{2}-4

3. Identify where the following parametric function intersects with its inverse.

x =4t

y =t^2-16

Answers:

1. The parameterized function is:

x_1 &=t \\y_1 &=t^2+t-4

The inverse is:

x_2 &=t^2+t-4 \\y_2 &=t

To find where these intersect, set  x_1=x_2  and  y_1=y_2  and solve.

t &=t^2+t-4 \\t^2 &=4 \\t &=\pm 2

You still need to actually calculate the points of intersection on the graph.  You can tell from the graph in Example C that there seem to be four points of intersection.  Since t can mean time, the question of intersection is more complicated than simply overlapping.  It means that the points are at the same x and y coordinate at the same time.  Note what the graphs look like when -1.8 < t < 1.8.

Note what the graphs look like t > 2.2  or t < -2.2

Notice how when these partial graphs are examined there is no intersection at anything besides t = \pm2 and the points (2, 2) and (-2, -2) While the paths of the graphs intersect in four places, they intersect at the same time only twice.

2. First check to see if the point (-2, 6) produces a matching time for the original function.

-2 &=t^2-10

8=t^2

\pm 2\sqrt{2}=t

6=\frac{t}{2}-4

20=t

The point does not live on the original function.  Now, you must check to see if it lives on the inverse.

6=t^2-10

\pm4=t

-2=\frac{t}{2}-6

4=\frac{t}{2}

8=t

 The point does not live on the inverse either.

3. x_1=4t; \  y_1=t^2-16 The inverse is:

x_2 &=t^2-16 \\y_2 &=4t

Solve for t  when  x_1=x_2  and  y_1=y_2.

4t &=t^2-16 \\0 &=t^2-4t-16 \\t &=\frac{4 \pm\sqrt{16-4 \cdot 1 \cdot (-16)}}{2}=\frac{4 \pm 4 \sqrt{5}}{2}=2 \pm 2 \sqrt{5}

The points that correspond to these two times are:

x &=4(2+2\sqrt{5}), \ y=(2+2\sqrt{5})^2 -16 \\x &=4(2-2\sqrt{5}), \ y=(2-2\sqrt{5})^2 -16 \\

Practice

Use the function x=t-4; \ y=t^2+2  for #1 - #3.

1. Find the inverse of the function.

2. Does the point (-2, 6) live on the function or its inverse?

3. Does the point (0, 1) live on the function or its inverse?

Use the relation x=t^2; \ y=4-t  for #4 - #6.

4. Find the inverse of the relation.

5. Does the point (4, 0) live on the relation or its inverse?

6. Does the point (0, 4) live on the relation or its inverse?

Use the function x=2t+1; \ y=t^2-3  for #7 - #9.

7. Find the inverse of the function.

8. Does the point (1, 5) live on the function or its inverse?

9. Does the point (9, 13) live on the function or its inverse?

Use the function x=3t+14; \ y=t^2-2t for #10 - #11.

10. Find the inverse of the function.

11. Identify where the parametric function intersects with its inverse.

Use the relation  x=t^2; \ y=4t-4  for #12 - #13.

12. Find the inverse of the relation.

13. Identify where the relation intersects with its inverse. 

14. Parameterize f(x)=x^2+x-6  and then graph the function and its inverse.

15. Parameterize  f(x)=x^2+3x+2  and then graph the function and its inverse.

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Parametric-Inverses.

Reviews

Please wait...
Please wait...

Original text