A cylinder gets taller at a rate of 3 inches per second, but the radius shrinks at a rate of 1 inch per second. How fast is the volume of the cylinder changing when the height is 20 inches and the radius is 10 inches?

Answer 1
Volume of a cylinder is #V = pir^2h# where
#r# is the radius and #h# is the height of the cylinder,
We are given some rates of change: #(dh)/(dt) = 3# #"in"#/#"s"# and #(dr)/(dt) = -1# #"in"#/#"s"#.
We seek #(dV)/dt# when #h=20# and #r = 10#.
Differentiate both sides of #V = pir^2h# with respect to #t#. We'll need the product rule on the right.
#d/dt(V) = d/dt(pir^2)h+pir^2 d/dt(h)#
#(dV)/dt = 2pir (dr)/dt h +pir^2(dh)/dt#

At the instant of interest,

#(dV)/dt = 2pi(10) (-1) (20) +pi(10)^2 (3) = -100pi# #"in"^3#/#"s"#
The volume is decreasing at a rate of #100pi# #"in"^3#/#"s"#
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Answer 2

Volume is decreasing at a rate of #100pi \ i n^3s^(-1)#,
or approximately #314\ i n^3s^(-1)#

Let us setup the following variables:

{

(r, "Radius of cylinder at time t","(in)"), (h, "Height of cylinder at time t","(in)"), (V, "Volume of cylinder at time t", "(in"^3")"), (t, "time", "(sec)") :} #

Using the standard formula for volume of a cylinder:

# V = pir^2h #
Where both #r# and #h# are functions of #t#. Implicitly differentiating wrt #t#, we have via the product rule:
# d/dt V = (pir^2)(d/dt h) + (d/dt pir^2)(h) #

Then by the chain rule, we have:

# (dV)/dt = (pir^2)( (dh)/dt d/(dh) h) + ( (dr)/dt d/(dr) pir^2)(h) # # \ \ \ \ \ \ = (pir^2)( (dh)/dt 1) + ( (dr)/dt 2pir)(h) # # \ \ \ \ \ \ = pir^2 (dh)/dt + 2pirh (dr)/dt #

And we are given that

# (dh)/dt=3# and #(dr)/dt = -1#

So we have:

# (dV)/dt = 3pir^2 - 2pirh #
When #h=20# and #r=10# we have:
# [(dV)/dt]_(h=20, r=10) = 3pi(100) - 2pi(10)(20) # # " " = 300pi - 400pi # # " " = -100pi #
Hence, the volume is decreasing at a rate of #100 \ i n^3s^(-1)#
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Answer 3

To find how fast the volume of the cylinder is changing when the height is 20 inches and the radius is 10 inches, we use the formula for the volume of a cylinder:

[ V = \pi r^2 h ]

We are given that the height is changing at a rate of 3 inches per second (( \frac{dh}{dt} = 3 )) and the radius is changing at a rate of -1 inch per second (( \frac{dr}{dt} = -1 )). We need to find the rate of change of volume (( \frac{dV}{dt} )) when ( h = 20 ) inches and ( r = 10 ) inches.

Using the given information and the chain rule, we differentiate the volume formula with respect to time:

[ \frac{dV}{dt} = \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) ]

Substitute the given values:

[ \frac{dV}{dt} = \pi (2(10)(20)(-1) + (10)^2 (3)) ]

[ \frac{dV}{dt} = \pi (-400 + 300) ]

[ \frac{dV}{dt} = -100\pi ]

Therefore, when the height is 20 inches and the radius is 10 inches, the volume of the cylinder is changing at a rate of ( -100\pi ) cubic inches per second.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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