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Here is the question verbatim:
The radius of a right-circular cylinder increases at a constant rate. Its altitude is a linear function of the radius and increases three times as fast as the radius. When the radius is 1 foot, the altitude is 6 feet. When the radius is 6 feet, the volume is increasing at a rate of 1 cubic feet per second. When the radius is 36 feet, the volume is increasing at a rate of n cubic feet per second, where n is an integer. Compute n.I set the problem up as follows:
The radius of a right-circular cylinder increases at a constant rate. Then let r be the radius and t be the number of seconds. This implies that \frac{dr}{dt}=c for some constant c.
Its altitude is a linear function of the radius Letting h be the altitude, we have h=ar+b for some constants a and b.
and increases three times as fast as the radius. Then \frac{dh}{dt}=3\frac{dr}{dt}=3c. Also, we have \frac{dh}{dr}=a and so by the chain rule, \frac{dh}{dr}\frac{dr}{dt}=\frac{dh}{dt} \implies a=2c as long as c\ne0, which we will have shortly.
When the radius is 1 foot, the altitude is 6 feet. So 6=2c+b.
When the radius is 6 feet, the volume is increasing at a rate of 1 cubic feet per second. Letting v be the volume, we have \frac{dv}{dt}=1. This is how we know that c\ne0, since \frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}.
Now we work to find n. We have 6=2c+b \implies b=6-2c \implies h=2cr+6-2c. The volume of a right-circular cylinder is given by v = \pi r^2 h =\pi r^2 (2cr+6-2c) = 2c\pi r^3+(6-2c)\pi r^2 and so \frac{dv}{dr}=6c\pi r^2 + 2(6-2c)\pi r. Using the chain rule, we have \frac{dv}{dt}=(6c\pi r^2 + 2(6-2c)\pi r)c and when r=6, 1=(6c\pi 6^2 + 2(6-2c)\pi 6)c which, simplified, is 1=192 \pi c^2+72 \pi c. From this alone, we can see that whatever c is, it is a very complicated number because it must remove the values of pi in the equation. Solving this using the quadratic equation will give us two possible values:
Solved with WolframAlpha
Using the positive one, since the radius is increasing, we should be able to compute \frac{dv}{dt} directly, however for r=36 we don't get an integer.
Final Solution from WolframAlpha
The book's answer is 33. Where did I go wrong?
The radius of a right-circular cylinder increases at a constant rate. Its altitude is a linear function of the radius and increases three times as fast as the radius. When the radius is 1 foot, the altitude is 6 feet. When the radius is 6 feet, the volume is increasing at a rate of 1 cubic feet per second. When the radius is 36 feet, the volume is increasing at a rate of n cubic feet per second, where n is an integer. Compute n.I set the problem up as follows:
The radius of a right-circular cylinder increases at a constant rate. Then let r be the radius and t be the number of seconds. This implies that \frac{dr}{dt}=c for some constant c.
Its altitude is a linear function of the radius Letting h be the altitude, we have h=ar+b for some constants a and b.
and increases three times as fast as the radius. Then \frac{dh}{dt}=3\frac{dr}{dt}=3c. Also, we have \frac{dh}{dr}=a and so by the chain rule, \frac{dh}{dr}\frac{dr}{dt}=\frac{dh}{dt} \implies a=2c as long as c\ne0, which we will have shortly.
When the radius is 1 foot, the altitude is 6 feet. So 6=2c+b.
When the radius is 6 feet, the volume is increasing at a rate of 1 cubic feet per second. Letting v be the volume, we have \frac{dv}{dt}=1. This is how we know that c\ne0, since \frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}.
Now we work to find n. We have 6=2c+b \implies b=6-2c \implies h=2cr+6-2c. The volume of a right-circular cylinder is given by v = \pi r^2 h =\pi r^2 (2cr+6-2c) = 2c\pi r^3+(6-2c)\pi r^2 and so \frac{dv}{dr}=6c\pi r^2 + 2(6-2c)\pi r. Using the chain rule, we have \frac{dv}{dt}=(6c\pi r^2 + 2(6-2c)\pi r)c and when r=6, 1=(6c\pi 6^2 + 2(6-2c)\pi 6)c which, simplified, is 1=192 \pi c^2+72 \pi c. From this alone, we can see that whatever c is, it is a very complicated number because it must remove the values of pi in the equation. Solving this using the quadratic equation will give us two possible values:
Solved with WolframAlpha
Using the positive one, since the radius is increasing, we should be able to compute \frac{dv}{dt} directly, however for r=36 we don't get an integer.
Final Solution from WolframAlpha
The book's answer is 33. Where did I go wrong?
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