Just a note of caution.
If this is a regular Calculus course, your questions will not become any easier than these.
1. V = (4/3)π r^3
dV/dt = 4πr^2 dr/dt
for your given data
dV/dt = 4π(64)(14)
= 3584π inches^3/min
SA = 4πr^2
dSA/dt = 8πr dr/dt
again , plug in your given
dSA/dt = 8π(8)(14)
= 896π inches^2/min
y = 4 - x^2
dy/dt = -2x dx/dt
but we need dy/dt = dx/dt, so
dx/dt = -2x dx/dt
divide by dx/dt
1 = -2x
x = -1/2
y = 4 - (-1/2)^2
= 4 - 1/4
= 15/4
the point is (-1/2, 15/4)
1. The radius of a sphere is expanding at a rate of 14 in/min.
a) Determine the rate at which the volume is changing when the radius is 8 in.
b) Determine the rate at which the surface area is changing when the radius is 8 in.
2. A particle moves along a path described by y=4-x^2. At what point along the curve are x and y changing at the same rate?
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