# Thread: Four tough related rate problems

1. ## Four tough related rate problems

Hey everyone, i am new here. I am in BC Calc at my school and we got a tough related rate problems worksheet over the weekend. I got the first two and the last one, but the these three have stumped me. Last year's problems weren't this hard. Can you help with whatever you can, please?

I took a picture of the problems. I know it's a lot, but if possible, just help with whatever you know.

Thanks so much!!!

2. So far, i have some work done.

3. I have dV/dt = k/r

V = (4/3)pi*r^3

dV/dt = 4pi*r^2 (dr/dt)

Thus, 4pi*r^2 (dr/dt) = k/r

I'm stuck here.

4. The volume is 8pi in^3

That's all i have for 4.

5. I don't' know how to relate the eqautions.

3. Hi there. I'm Nowegian and don't know what reciprocal means, but if the increase of r (referring to task 3) is constant (r(t) is linear), then you can find the rate of increase of the radius by $\frac{r_{t_1}-r_{t_0}}{t_1-t_0}$. If we define $\Delta t=t_1-t_0$ we then get $\frac{r_{t_0+\Delta t}-r_{t_0}}{\Delta t}$, which is the definition of the derivate. (For linear functions anyway. For other functions let $\Delta t$ go to zero.) You know that $r_{t_1}=2$ and $r_{t_0}=1$ and $t_1-t_0=15$. You can use this to find the rate of increase of the radius, then apply it to problem B.

4. #5:

Circumference is $C=2{\pi}r$

and we know that $\frac{dC}{dt}=6$

Therefore, $\frac{dC}{dt}=2{\pi}\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{3}{\pi}$

The perimeter of the square is P=8r.

So, $\frac{dP}{dt}=8\frac{dr}{dt}$

$\frac{dP}{dt}=8(\frac{3}{\pi})=\frac{24}{\pi}$

For part b, you need the difference between the area of the circle and the square.

Since the area of the circle is ${\pi}r^{2}=25{\pi}$

r=5. Then a side length of the square is 10.

Continue on with this part?.

Can you use this to solve other ones now?.

5. Hello, SuperTyphoon!

5. A circle is inscribed in a square. The circumference of the circle is increasing at a constant
rate of 6 in/sec. As the circle expands, the square expands to maintain tangency.

A. Find the rate at which the perimeter of the square is increasing.
Code:
      *-------*-*-*-------*
|   *           *   |
| *               * |
|*                 *|
|                   |
*                   *
2r *         * - - - - *
*              r    *
|                   |
|*                 *|
| *               * |
|   *           *   |
*-------*-*-*-------*
2r

The radius of the circle is $r.$
The side of the square is $2r.$

The circumference of the circle is: . $C \:=\:2\pi r$

Differentiate with respect to time: . $\frac{dC}{dt} \:=\:2\pi\!\cdot\!\frac{dr}{dt}$

We are told that $\frac{dC}{dt} = 6$

So we have: . $6 \:=\:2\pi\!\cdot\!\frac{dr}{dt} \quad\Rightarrow\quad\frac{dr}{dt} \:=\:\frac{3}{\pi}$ in/sec. .[1]

The perimeter of the square is: . $P \:=\:4(2r) \:=\:8r$

Differentiate with respect to time: . $\frac{dP}{dt} \:=\:8\!\cdot\!\frac{dr}{dt}$ .[2]

Substitute [1] into [2]: . $\frac{dP}{dt} \:=\:8\left(\frac{3}{\pi}\right) \:=\:\boxed{\frac{24}{\pi}\text{ in/sec}}$

B. When the area of the circle is 25π inē,
find the rate on increase of the area between the circle and the square.

The area of the circle is: . $A_c \:=\:\pi r^2$

If the area is $25\pi$ inē, we have: . $25\pi \:=\:\pi r^2\quad\Rightarrow\quad r \:=\:5$ .[3]

The area of the square is: . $A_s \:=\:(2r)^2 \:=\:4r^2$

The difference of the areas is: . $D \:=\:A_s - A_c \:=\:4r^2 - \pi r^2\quad\Rightarrow\quad D \:=\:(4-\pi)r^2$

Differentiate with respect to time: . $\frac{dD}{dt} \:=\:2(4-\pi)r\!\cdot\!\frac{dr}{dt}$ .[4]

Substitute [1] and [3] into [4]: . $\frac{dD}{dt} \:=\:2(4-\pi) (5)\left(\frac{3}{\pi}\right) \:=\:\boxed{\frac{30(4-\pi)}{\pi}\text{ in}^2\text{/sec}}$

6. OMG Thank you guys so much!!! You don't know how much stress this has saved me. I understand it very well the way you write them too! Now i just need number 4, but still, thanks for all that.