MHB How Do You Solve Related Rates Problems Involving Melting Ice?

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The discussion focuses on solving related rates problems involving melting ice, specifically how to derive the rate of change of the radius of a sphere as it melts. Part A confirms the correct application of the volume formula and the differentiation process. In Part B, the main goal is to determine the time it takes for the sphere to completely melt, leading to the integration of the rate equation. The solution involves setting the radius to zero and solving for time, resulting in the formula t = r0/k, where r0 is the initial radius and k is a constant related to the melting rate. The conversation emphasizes the importance of initial conditions and proper integration techniques to arrive at the final answer.
tomc612
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Hi,
need some help on the following question.

Just want to check on part a on the followingv=4/3\pi.r^3

dv = 4\pi.r^2 dr

dv/dt = 4\pi.r^2 dr/dt

dr/dt = (dv/dt)/ 4\pi.r^2

dr/dt = (-KA)/4\pi.r^2

dr/dt= -K

part B need some help

Thanks

TomView attachment 6214
 

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tomc612 said:
Hi,
need some help on the following question.

Just want to check on part a on the followingv=4/3\pi.r^3

dv = 4\pi.r^2 dr

dv/dt = 4\pi.r^2 dr/dt

dr/dt = (dv/dt)/ 4\pi.r^2

dr/dt = (-KA)/4\pi.r^2

dr/dt= -K

part B need some help

Thanks

Tom

You've done part a) correctly.

As for part b, you want to know how long it takes for the sphere to melt entirely, in other words, for r to become 0.

So integrate $\displaystyle \begin{align*} \frac{\mathrm{d}r}{\mathrm{d}t} = -k \end{align*}$ with respect to t to get r in terms of t, and then solve for t where $\displaystyle \begin{align*} r = 0 \end{align*}$.
 
So,
dr/dt = -k

dr = -k.dt

intergral dr = integral -k.dt

r = -kt + c

0 = -kt +c

Still not sure that's it, or should it be that the integral of -k.dt is -1/2k^2t

Thanks
 
tomc612 said:
So,
dr/dt = -k

dr = -k.dt

intergral dr = integral -k.dt

r = -kt + c

0 = -kt +c

Still not sure that's it, or should it be that the integral of -k.dt is -1/2k^2t

Thanks
You have not yet answered the question! What is t when r= 0? Of course, that will depend upon what "c" is. You were given "k" as part of the problem when you were given \frac{dV}{dt}= -kA. To determine "c" use the fact that "the ice sphere has initial radius r_0 when t= 0".
 
so do we need to bring the Volume formula back into represent V and R
if V =4/3pi.r^3

then r = (3V/4pi)^1/3r0=-kt +c

(3V/4Pi)^1/3 =-kt+c

is that the right path?

Thanks
 
For part (b), we are to solve the IVP:

$$\d{r}{t}=-k$$ where $$r(0)=r_0$$

Integrate w.r.t $t$, using the given boundaries and switch dummy variables of integration:

$$\int_{r_0}^{r(t)}\,da=-k\int_0^t\,db$$

Apply the FTOC:

$$r(t)-r_0=-kt$$

Solve for $t$:

$$t=\frac{r_0-r(t)}{k}$$

To determine how long it takes the ice to melt away, we set $r(t)=0$:

$$t=\frac{r_0}{k}$$
 

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