MATHEMATICS COLLEGE

The height of a cylinder is decreasing at a constant rate of 1 centimeters per minute, and the volume is decreasing at a rate of 2341 cubic centimeters per minute. At the instant when the height of the cylinder is 1010 centimeters and the volume is 577 cubic centimeters, what is the rate of change of the radius? The volume of a cylinder can be found with the equation V=pi r^2 h. Round your answer to three decimal places.

Answers

Answer 1

Answer:

Let's organize our given information.

$(dh)/(dt)=-1$

$(dV)/(dt)=-2341$

$(dr)/(dt)=?, h=1010, V=577$

$V=\pi r^2h$

What I'd like to do is solve for $r$ first, it'll help later.

$577=\pi r^2(1010)$

$r=\sqrt{(577)/(1010\pi)}$

$r=0.426$

Now, we can differentiate our formula for the volume of a cylinder to find the rate of change of the radius.

$V=\pi r^2h$

Use the product rule of differentiation.

$(dV)/(dt)=2\pi rh(dr)/(dt)+\pi r^2(dh)/(dt)$

Plug in the information we are given.

$-2341=2\pi (0.426)(1010)(dr)/(dt)+\pi (0.426)^2(-1)$

Solve for $(dr)/(dt)$ .

$-2341=860.52\pi(dr)/(dt)+(0.426)^2\pi$

$-2341.57=860.52\pi(dr)/(dt)$

$(dr)/(dt)=-0.866$

The rate of change of the radius is -0.866 centimeters per minute.

Hope this helps!

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Answers

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Answers

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Step-by-step explanation:

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Answers

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Answers

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Answers

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Answers

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