Related rates and differential

[tomc612]

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

 

[Prove It]

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 $\displaystyle \begin{align*} r = 0 \end{align*}$.

 

[tomc612]

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

 

[HallsofIvy]

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 [tex]\frac{dV}{dt}= -kA[/tex]. To determine "c" use the fact that "the ice sphere has initial radius [tex]r_0[/tex] when [tex]t= 0[/tex]".

 

[tomc612]

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

 

[MarkFL]

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

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

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

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

Apply the FTOC:

\(\displaystyle r(t)-r_0=-kt\)

Solve for $t$:

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

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

\(\displaystyle t=\frac{r_0}{k}\)