A balloon is being filled at a rate of 4000 cm^3/min. How fast is the surface area of the ballon increasing when the radius is 125 cm? V=4/3pir^3, SA=4pir^2

Answer 1

V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
when r = 125
4000 = 4π(125^2) dr/dt
dr/dt = 4000/(62500π) =8/(125π)

SA = 4πr^2
d(SA)/dt = 8πr dr/dt
= 8π(125)(8/125π) = 64

check my arithmetic and put in the units. (64 cm^2)

Answer 2

Find the equation of the tangent line to the curve y=e^2x + 4x at the point (0,1)

Answer 3

the tangent line at any point (x,y) has slope y'

y = e^2x + 4x
y' = 2e^2x + 4

y'(0) = 6

so, now you have a point, (0,1), and a slope, 6.

using the point-slope form for a line,

(y-1)/(x-0) = 6
y-1 = 6x
y = 6x+1