Question

Let $V$ be the volume of a cylinder having height $h$ and radius $r,$ and assume that $h$ and $r$ vary with time. (a) How are $d V / d \tau, d h / d t,$ and $d r / d t$ related? (b) At a certain instant, the height is 6 in and increasing at 1 in/s, while the radius is 10 in and decreasing at 1 in $/ \mathrm{s}$ How fast is the volume changing at that instant? Is the volume increasing or decreasing at that instant?

          Let $V$ be the volume of a cylinder having height $h$ and radius
$r,$ and assume that $h$ and $r$ vary with time.
(a) How are $d V / d \tau, d h / d t,$ and $d r / d t$ related?
(b) At a certain instant, the height is 6 in and increasing at 1 in/s, while the radius is 10 in and decreasing at 1 in $/ \mathrm{s}$ How fast is the volume changing at that instant? Is the volume increasing or decreasing at that instant?
        
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Added by Michael B.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Let $V$ be the volume of a cylinder having height $h$ and radius $r,$ and assume that $h$ and $r$ vary with time. (a) How are $d V / d \tau, d h / d t,$ and $d r / d t$ related? (b) At a certain instant, the height is 6 in and increasing at 1 in/s, while the radius is 10 in and decreasing at 1 in $/ \mathrm{s}$ How fast is the volume changing at that instant? Is the volume increasing or decreasing at that instant?
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Transcript

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00:01 Hi there, so for this problem, we are told to denote the volume of a cylinder as capital b.
00:08 So we know that the volume of a cylinder is pi times its radius square times the height.
00:15 So with that said, for part a of this problem, we are asked about how are the differential in the volume, the differential in the height, and the differential in the radius related.
00:27 So what we are going to do is to take the volume that we are given, and thereby both sides with respect to time.
00:35 So for the left side, we obtain just simply the derivative of the volume with respect to time.
00:40 And for the right side, taking out the constant, which is pi, we have the derivative of the product between the radius and the height.
00:48 So we can apply in here the derivative of a product.
00:54 So that will be pi.
00:55 And the first derivative is two times the radius times the high, times the rate of change of the radius could respect the time.
01:09 This plus the radius square times the rate of change of the high could respect the time.
01:18 So that is the relationship between the differentials.
01:22 That's a solution for part a of this problem...
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