A right circular cylinder is changing shape. The radius is decreasing at the rate of 2 inches/second while its height is increasing at the rate of 5 inches/second. When the radius is 4 inches and the height is 6 inches, how fast is the
a) volume changing (V=(pi)r^2h)
b) surface area changing (S=2(pi)rh)
dV/dt = πr^2 dh/dt + 2πrh dr/dt
plug in the given values
SA = 2πrh , are you not considering the ends ?
differentiate with respect to t using the product rule like I did for the volume, then plug in the values
To find how fast the volume and surface area are changing, we can use the given rates of change and the formulas for volume and surface area of a cylinder.
a) To find how fast the volume is changing, we need to differentiate the volume formula with respect to time:
V = πr^2h
We have given that the radius is decreasing at a rate of 2 inches/second, so dr/dt = -2 in/s (negative because it's decreasing). Also, the height is increasing at a rate of 5 inches/second, so dh/dt = 5 in/s. We want to find dV/dt, the rate of change of the volume with respect to time:
dV/dt = d(πr^2h)/dt
Using the product rule of differentiation, we can write this as:
dV/dt = 2πrh(dr/dt) + πr^2(dh/dt)
Plugging in the given values:
r = 4 in
h = 6 in
dr/dt = -2 in/s
dh/dt = 5 in/s
we can calculate dV/dt:
dV/dt = 2π(4)(6)(-2) + π(4^2)(5)
dV/dt = -48π - 80π
dV/dt = -128π in^3/s
So, the volume is changing at a rate of -128π cubic inches per second.
b) To find how fast the surface area is changing, we need to differentiate the surface area formula with respect to time:
S = 2πrh
Again, we can use the given rates of change to find dS/dt, the rate of change of the surface area with respect to time:
dS/dt = d(2πrh)/dt
Using the product rule of differentiation:
dS/dt = 2πh(dr/dt) + 2πr(dh/dt)
Plugging in the given values:
r = 4 in
h = 6 in
dr/dt = -2 in/s
dh/dt = 5 in/s
we can calculate dS/dt:
dS/dt = 2π(6)(-2) + 2π(4)(5)
dS/dt = -24π + 40π
dS/dt = 16π in^2/s
So, the surface area is changing at a rate of 16π square inches per second.
To find how fast the volume and surface area of the cylinder are changing, we need to use the chain rule from calculus.
a) To find the rate at which the volume is changing, we'll take the derivative of the volume formula with respect to time, t:
V = πr^2h
Differentiating both sides with respect to t:
dV/dt = d/dt(πr^2h)
Using the chain rule, we have:
dV/dt = π(2rh(dr/dt) + r^2(dh/dt))
Now we can substitute the given values:
r = 4 inches, dr/dt = -2 inches/second (since the radius is decreasing), h = 6 inches, and dh/dt = 5 inches/second.
Plugging these values into the derivative equation, we get:
dV/dt = π(2(4)(6)(-2) + 4^2(5))
Simplifying further:
dV/dt = π(-96 + 80)
dV/dt = -16π cubic inches/second
Therefore, the volume is decreasing at a rate of 16π cubic inches per second.
b) To find the rate at which the surface area is changing, we'll take the derivative of the surface area formula with respect to time, t:
S = 2πrh
Differentiating both sides with respect to t:
dS/dt = d/dt(2πrh)
Using the chain rule, we have:
dS/dt = 2π(dh/dt)(r) + 2π(dr/dt)(h)
Substituting the given values:
r = 4 inches, dr/dt = -2 inches/second, h = 6 inches, and dh/dt = 5 inches/second.
Plugging these values into the derivative equation, we get:
dS/dt = 2π(5)(4) + 2π(-2)(6)
Simplifying further:
dS/dt = 40π - 24π
dS/dt = 16π square inches/second
Therefore, the surface area is changing at a rate of 16π square inches per second.