How Do You Solve Related Rates Problems Involving Melting Ice?

In summary, to find the time it takes for the ice sphere to melt completely, we use the formula t = r_0/k, where r_0 is the initial radius of the sphere and k is a constant given in the problem. This is found by solving the initial value problem, where r(t) represents the radius of the sphere at time t and the derivative of r with respect to t is equal to -k.
  • #1
tomc612
17
0
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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  • #2
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 $\displaystyle \begin{align*} r = 0 \end{align*}$.
 
  • #3
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
 
  • #4
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 [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]".
 
  • #5
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
 
  • #6
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 $r(t)=0$:

\(\displaystyle t=\frac{r_0}{k}\)
 

FAQ: How Do You Solve Related Rates Problems Involving Melting Ice?

1. What is the difference between related rates and differential equations?

Related rates involve finding the rate of change of one variable with respect to another, while differential equations involve finding the relation between a function and its derivatives. Related rates are typically used to solve specific problems, while differential equations are used to model systems or phenomena.

2. What are the steps to solving a related rates problem?

The first step is to identify the variables involved and their rates of change. Then, use equations and geometric relationships to express the rates of change in terms of the variables. Next, differentiate both sides with respect to time and solve for the desired rate of change. Finally, substitute the known values and solve for the unknown rate of change.

3. How are related rates and derivatives related?

Related rates involve taking the derivative of a function with respect to time, while differential equations involve finding the derivatives of functions in relation to each other. In both cases, derivatives are used to determine the rate of change of a variable.

4. Can related rates be applied to real-world situations?

Yes, related rates are commonly used in physics, engineering, and other scientific fields to solve problems involving changing quantities. For example, related rates can be used to determine the speed of a car, the rate at which a balloon is inflating, or the rate at which water is flowing into a tank.

5. What are some common applications of differential equations?

Differential equations are widely used in physics, engineering, economics, and other fields to model systems and phenomena. Examples include the motion of a pendulum, population growth, and heat transfer in a system. Differential equations can also be used to solve optimization problems and predict future behavior of systems.

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