Question

Using calculus, find the point on the circle x² + y² = R² that is closest to the point (R, R). What about the farthest one? Sketch the problem. Hint: extremize the square of the distance.

Answer 1

To find the point on the circle x² + y² = R² that is closest to the point (R, R), use calculus to extremize the square of the distance between the two points. The closest point is (R, R), while the farthest point is (-R, -R).

Step-by-step explanation:

To find the point on the circle x² + y² = R² that is closest to the point (R, R), we need to extremize the square of the distance between the two points. Let's denote the distance between the two points as D.

D² = (x - R)² + (y - R)²

Now, we can use calculus to find the minimum value of D². Taking the derivative of D² with respect to x and y, and setting them equal to zero, we get:

2(x - R) = 0 ---> x = R
2(y - R) = 0 ---> y = R

So, the point on the circle closest to (R, R) is (R, R).

The farthest point on the circle would be the diametrically opposite point. So, if we assume the circle to be centered at the origin (0, 0), then the farthest point would be (-R, -R).