Respuesta :
Distance between the origin and (3, 9) is [tex] \sqrt{ 3^{2}+ 9^{2} } [/tex]
=[tex] \sqrt{90} [/tex]
=3[tex] \sqrt{10} [/tex] square units
=[tex] \sqrt{90} [/tex]
=3[tex] \sqrt{10} [/tex] square units
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
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