Answered: An inflatable pumpkin and a ghost are…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

An inflatable pumpkin and a ghost are connected with an airpump. As the pumpkin deflates, the ghost inflates, then the process reverses. Assume that no air is leaking out, only moving between the pumpkin and the ghost. In addition, assume that the approximate shape of the pumpkin is a sphere, and the approximate shape of the ghost is a cone. The pumpkin’s radius is decreasing by 0.5 inches per second. Assume that the ghost’s height is always 10 times the radius of its base. Find the rate at which the ghost’s height is increasing at the moment when the diameter of the pumpkin is 40 inches, and the radius of the ghost’s base is 10 inches.

Expert Solution

Step 1

Step:-1

Volume of Sphere = $\frac{4}{3} {πR}^{3};$ R is radius of sphere

Volume of cone = $\frac{{πr}^{2} h}{3}$ ; r is radius of base and h is height.

Given that one pumpkin (sphere) and one ghost (cone) both are connected by air-pump. Also, the pumpkin inflates, then the ghost inflates, and reverse

that is, volume of air is same for both and

Volume of air = Volume of pumpkin = volume of ghost

$\Rightarrow$ Volume of pumpkin = volume of ghost ----- (1)

Let the volume of pumpkin is $v_{p}$ and volume of ghost is $v_{g}$

Also, pumpkin is sphere, so $v_{p}$ = $\frac{4}{3} {πR}^{3}$

and

$\frac{d v_{p}}{d t} = \frac{4}{3} π \times 3 \times R^{2} \times \frac{dR}{dt} = 4 {πR}^{2} \frac{dR}{dt} \Rightarrow \frac{{dv}_{p}}{dt} = 4 {πR}^{2} \frac{dR}{dt} - - - - (2)$

Given that Ghost is cone, so volume is $v_{g} = \frac{{πr}^{2} h}{3}$

And given that h=10 r

So, volume of ghost is

$v_{g} = \frac{{πr}^{2} h}{3} = \frac{1}{3} π (\frac{h^{2}}{10^{2}}) h = \frac{{πh}^{3}}{3 \times 100} v_{g} = \frac{{πh}^{3}}{3 \times 100}$

And $\frac{d v_{g}}{d t} = \frac{{πh}^{2}}{100} \frac{d h}{d t}$ --- (3)

Step:-2

By (1), we have

$\Rightarrow v_{p} = v_{g}$

Differentiating it with respect to time t, to find rate of change. We get

$\frac{d v_{p}}{d t} = \frac{d v_{g}}{d t}$

by (2) and (3),

$4 {πR}^{2} \frac{d R}{d t} = \frac{{πh}^{2}}{100} \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = 400 \times {(\frac{R}{h})}^{2} \frac{d R}{d t} - - - - (4)$

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