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Problem 36 Hard Difficulty

A faucet is filling a hemispherical basin of diameter $60 \mathrm{~cm}$ with water at a rate of $2 \mathrm{~L} / \mathrm{min}$. Find the rate at which the water is rising in the basin when it is half full. [Use the following facts: $1 \mathrm{~L}$ is $1000 \mathrm{~cm}^{3}$. The volume of the portion of a sphere with radius $r$ from the bottom to a height $h$ is $V=\pi\left(r h^{2}-\frac{1}{3} h^{3}\right),$ as we will show in Chapter $6 .$

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hemispherical basin. So hemispherical means half a sphere and its diameter. 60 centimeters that's being filled at a rate of two liters per minute. Find the rate the water is rising when the basin is half full. Okay, so first we need to draw a picture. So let's just draw side view half a circle and then put some water in there at the bottom. Okay. What we know so far is that the radius of the bowl is 37 30 centimeters, that's all. Okay, the water is going in at a rate of two liters per minute. So what that is doing is changing the volume of the water. So that's D V D t. Let's make that one blue D v D T rate of change of the volume with respected time. Two leaders per minute Find the right. The water is rising. So we're trying to find let's call it H here from the bottom of the basin to the top of the water. So find a d h d t when the basin is half full. So when it's half full, then h will be 15. So when H equals 15 centimeters okay, so we know the volume of a sphere is four thirds pi r cubed. But we don't have a whole sphere here. We're only feeling got a part of it filled up. So we need a special formula which they gave us right here. The volume of a portion of the sphere is pi. Times are h squared, minus one third h cute. So that's what we're gonna take the derivative of. Oh, and that are right. There is not the radius of the water but the radius of the sphere. Okay, First, we got to change this DVD t into Centimeters Cube. So we also are given this little bit of information. One leader is 1000 centimeters cube, so D V D t 2000 centimeters huge per minute. All right, so v is pi. Times are h squared minus one third h cube. So we're gonna take the drift, Give with respect to time So derivative of B Do you really see plays? A constant just comes along Oh, I forgot to put the radius and its 30 derivative of age square to age times the derivative of age D h d t minus one third time's derivative of H cube three h squared D h d t. All right now we gotta do is put the numbers in and simplify. We'll be all done 2000 equals pi times 60 times 15 d h d t minus 15 squared d h d t. So let's divide the pi over here. 2000 Divide by pi equals. So we have 60 times 15 minus 15 times. 15, um 60 is four times 15. I'm just doing it this way. So I don't have to get my calculator out, so I'm gonna factor of the 15 times 15 out I get four minus one there. So 15 times 15, that's to 25 times three 6 75. Yeah. All right. The H d t. So 2000 over 6. 75 pi equals d h d t. Okay, so DVT was in centimeters cubed. We divided by centimeter. Centimeters. So Senator used to keep per minute. So this is in centimeters per minute. Okay, We can reduce the 2000 and the 6 75 by Well, let's just do five. So 401 35 80. And I'm doing five again. I have 27 Okay, So D h d t 80/27 pi centimeters per minutes.

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James Stewart

Calculus: Early Transcendentals

Heather Zimmers

Samuel Hannah

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Heather Zimmers

Samuel Hannah

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Heather Zimmers

Samuel Hannah

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals