Problem: Consider (1) the parabola y=3-1/10 x^2 and (2) the upper half of the circle centered at (20, 0) with radius of 10. Find the points on the parabola where the tangent line is also tangent to the upper half of the circle. (You can find these points in exact form in Mathematica, but they are so messy that you will want to approximate them.) Make a single, final graph of the parabola, semicircle, and tangent line(s).

Question

Problem: Consider (1) the parabola y=3-1/10 x^2 and (2) the upper half of the circle centered at (20, 0) with radius of 10. Find the points on the parabola where the tangent line is also tangent to the upper half of the circle. (You can find these points in exact form in Mathematica, but they are so messy that you will want to approximate them.) Make a single, final graph of the parabola, semicircle, and tangent line(s).

Hint to plot a circle that really looks like a circle: In Show, set PlotRange->{{xmin, xmax},{ymin,ymax}} with the distance between xmin and xmax the same as the distance from ymin to ymax. Also set AspectRatio -> 1.

Answer 1

at x=h, the tangent line to the parabola is

y - (3-h^2/10) = (-h/5)(x-h)
y = -h/5 x + 3+h^2/10

So, find where the line intersects the circle, and find h so that there is only one solution.

It might be useful to pick a value of x close to what you expect, and then use an iterative method to close in on the needed slope for the line.

http://www.wolframalpha.com/input/?i=plot+y%3D3-x%5E2%2F10,+(x-20)%5E2%2By%5E2+%3D+100,+y%3D0,+x%3D0

Answer 2

To find the points on the parabola where the tangent line is also tangent to the upper half of the circle, we need to start by finding the equation of the tangent line and then determine the points of intersection between the tangent line and the parabola.

Let's begin by finding the equation of the parabola:

1) The equation of the parabola is given as y = 3 - (1/10)x^2.

Next, let's find the equation of the upper half of the circle centered at (20, 0) with a radius of 10:

2) The equation of the circle can be written as (x - 20)^2 + y^2 = 10^2.

Now we can proceed with finding the points of intersection between the parabola and the upper half of the circle.

To find the tangent lines to both the parabola and the circle, we can take the derivative of the equations and set them equal to each other, since the slope of the tangent line is the same for both curves at the point of tangency.

Differentiating the equation of the parabola, we get:

dy/dx = (-1/5)x.

Differentiating the equation of the circle, we get:

dy/dx = (20 - x)/y.

Setting the derivatives equal to each other, we have:

(-1/5)x = (20 - x)/y.

We can solve this equation for x to find the x-coordinate(s) of the points of intersection. Substituting the value of y from the equation of the parabola, we have:

(-1/5)x = (20 - x)/(3 - (1/10)x^2).

Simplifying this equation will give us approximate values for the x-coordinate(s).

Now let's plot the parabola, the upper half of the circle, and the tangent lines using this solution:

1. Define the equations for the parabola, the upper half of the circle, and the tangent lines.
2. Calculate the approximate x-coordinate(s) of the points of intersection.
3. Use the calculated x-coordinate(s) to find the corresponding y-coordinate(s) on the parabola.
4. Plot the parabola, the upper half of the circle, and the tangent lines on a graph using Mathematica.
5. Ensure that the plot shows a circle that resembles a circle by setting PlotRange and AspectRatio appropriately.

By following these steps, you will be able to make a final graph that shows the parabola, semicircle, and the tangent line(s) at the points of tangency.