Question

A bug is moving along the right side of the parabola y=x^2 at a rate such that its distance from the origin is increasing at 7 cm/min. At what rates are the​ x- and​ y-coordinates of the bug increasing when the bug is at the point (3, 9)​?

Answer 1

Distance between the origin and (3, 9) is
`\sqrt{ 3^(2)+ 9^(2) }`
=
`√(90)`
=3
`√(10)` square units

Answer 2

Let D be the distance of the bug from the origin.
D = √(x² + y²)
dD/dx = x/√(x² + y²)
dD/dy = y/√(x² + y²)
dD/dt = 7 cm/s
We need dx/dt and dy/dt. We will use the chain rule
dx/dt = dx/dD * dD/dt
= √(x² + y²)/x * 7 = [7√(x² + y²)]/x
Applying the same method for y,
dy/dt = [7√(x² + y²)]/y
Now, we substitute the values of x and y to get the rates.
dx/dt = [7√(x² + y²)]/x = (7√90)/3 = 22.1
dx/dy = [7√(x² + y²)]/y] = (7√90)/9 = 7.38