A circle has center that the origin and radius r. What is r when the circle is tangent e^(-x^2) (but no intersection with e^(-x^2))? explain.... I can graph it and estimate it, but I don't know how to do this by hand.

Question

A circle has center that the origin and radius r. What is r when the circle is tangent e^(-x^2) (but no intersection with e^(-x^2))? explain.... I can graph it and estimate it, but I don't know how to do this by hand.

Answer 1

Not sure what tools you have to use, but you might try to minimize the distance from (0,0) to e^(-x^2). That is

z^2 = x^2 + e^(-2x^2)
2z z' = 2x - 4x e^(-2x^2)
Since z will never be zero, you just need

x = 2xe^(-2x^2)
e^(-2x^2) = 1/2
x^2 = ln2/2
So, y^2 = e^(-ln2/2) = 1/2

x^2+y^2 = ln2/2 + 1/2

See the graphs at

http://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%3Dln2%2F2+%2B1%2F2,+y+%3D+e%5E(-x%5E2)

Answer 2

To determine the radius, r, of a circle that is tangent to the curve e^(-x^2) without intersecting it, we need to analyze the properties of the circle and the function.

First, let's consider the equation of the circle with center at the origin (0,0) and radius r:

x^2 + y^2 = r^2

Next, we need to examine the curve e^(-x^2). It is a symmetric function with its highest point at x = 0 and decreases rapidly as x moves away from zero in either direction. Since the circle is tangent to the curve, it would touch the curve at a point with the same x-coordinate as the highest point of the curve (x = 0).

So, to find the radius, we need to determine the y-coordinate of the highest point of the curve e^(-x^2).

To find this point, we can take the derivative of the function e^(-x^2) with respect to x and set it equal to zero:

d/dx (e^(-x^2)) = 0

Using the chain rule, we differentiate e^(-x^2) to get:

-2x * e^(-x^2) = 0

Since e^(-x^2) is never equal to zero, the only way for the derivative to be zero is for -2x to be zero. Therefore, x = 0.

Substituting x = 0 back into the function, we obtain:

e^(-0^2) = e^0 = 1

Therefore, the y-coordinate of the highest point on the curve is y = 1.

Since the circle is tangent to the curve at this point, it implies that the y-coordinate of the highest point of the circle is also y = 1.

Now we can substitute this point into the equation of the circle:

0^2 + 1^2 = r^2

1 = r^2

Taking the square root of both sides, we find:

r = 1

Hence, the radius, r, of the circle that is tangent to the curve e^(-x^2) (but no intersection) is 1.

A circle has center that the origin and radius r. What is r when the circle is tangent e^(-x^2) (but no intersection with e^(-x^2))? explain.... I can graph it and estimate it, but I don't know how to do this by hand.

Not sure what tools you have to use, but you might try to minimize the distance from (0,0) to e^(-x^2). That is

Thank you

To determine the radius, r, of a circle that is tangent to the curve e^(-x^2) without intersecting it, we need to analyze the properties of the circle and the function.