# Math Help - [SOLVED] 3 problems

1. ## [SOLVED] 3 problems

If anyone can help me with the answers and work to these problems that would be great.

1. The surface area of a balloon for a parade is increasing at a constant rate of 10.8 square yards per second. Find the rate of change of the volume of the balloon when the diameter is 350 feet

2. Sand pouring from a chute forms a conical pile whose height is always equal to twice the diameter. If the height increases at a constant rate of 8 feet per minute what rate is the sand pouring from the chute when the pile is 50 feet high.

3. Find the area of the region bound by the following graph

f(x)= -x^2+4x+2
g(x)= x+2

2. 2. Sand pouring from a chute forms a conical pile whose height is always equal to twice the diameter. If the height increases at a constant rate of 8 feet per minute what rate is the sand pouring from the chute when the pile is 50 feet high.
We are given dh/dt=8, h/4=r, we want dV/dt when h=50.

Volume of cone is $\frac{\pi}{3}r^{2}h$

Using our sub, r=h/4:

$V=\frac{\pi}{3}(\frac{h}{4})^{2}h=\frac{{\pi}h^{3} }{48}$

Now, it is in terms of h alone. We are given dh/dt.

So, differentiating: $\frac{dV}{dt}=\frac{{\pi}h^{2}}{16}\cdot\frac{dh}{ dt}$

Now, by using our givens:

$\frac{dV}{dt}=\frac{{\pi}(50)^{2}}{16}\cdot{8}=125 0{\pi}\approx{3927} \;\ ft^{3}/min$

That's stacking up pretty fast. But, then again, the pile is growing 8 feet per minute, so dV/dt would have to be high.

3. Originally Posted by bstewart09
If anyone can help me with the answers and work to these problems that would be great.

1. The surface area of a balloon for a parade is increasing at a constant rate of 10.8 square yards per second. Find the rate of change of the volume of the balloon when the diameter is 350 feet

2. Sand pouring from a chute forms a conical pile whose height is always equal to twice the diameter. If the height increases at a constant rate of 8 feet per minute what rate is the sand pouring from the chute when the pile is 50 feet high.

3. Find the area of the region bound by the following graph

f(x)= -x^2+4x+2
g(x)= x+2

For number three setting $f(x)=g(x)$ we get x=0 and x=3

Now we must test and element of this interval in f(x) and g(x)

So $f(1)=5$ and $g(1)=3$

$\therefore,f(x)>g(x)\forall{x}\in(1,3)$

So the area is equal to $\int_0^{3}[f(x)-g(x)]dx=\int_0^{3}(-x^2+4x-2)-(x+2)dx=\int_0^{3}-x^2+3xdx=\bigg[\frac{-x^3}{3}+\frac{3x^2}{2}\bigg]\bigg|_0^{3}$ $=\frac{9}{2}\text{ }\text{square units}$

4. Thanks for those two answer, now just to get the first one answered.

P.S. sorry bout the bump, didn't know it was against the rules

5. Originally Posted by bstewart09
Thanks for those two answer, now just to get the first one answered.

P.S. sorry bout the bump, didn't know it was against the rules
$V_{sphere}=\frac{4}{3}\pi{r^3}$

and $SA_{sphere}=4\pi{r^2}$

So subbing this into we have $V=(4\pi{r^2})\cdot\frac{r}{3}\Rightarrow{V=SA\cdot \frac{r}{3}}$

Differenatiating we get

$V'=(SA)'\cdot\frac{r}{3}+\frac{SA}{3}$

Since this is when $d=350\Rightarrow{r=\frac{350}{2}=175ft=\frac{175}{ 3}\text{yards}}$

and since $r=\frac{175}{3}$ we calculate $SA=4\pi{r^2}=4\pi(175)^2=\frac{122500\pi}{9}$

and finally we know that at this point $(SA)'=10.8\text{ }\text{square yards}$
so
So $V'=\bigg[(10.8)\cdot\frac{175}{3}+\frac{122500\pi}{27}\bigg]\text{ }\text{cubic yards per second}$

I leave teh final calculation up to you...I hope I didnt make computation error..that seemed messy

6. Be careful of the units. 10.8 yards equals 32.4 feet.

$V=\frac{4}{3}{\pi}r^{3}$

$S=4{\pi}r^{2}$

But $\frac{dS}{dt}=8{\pi}r\frac{dr}{dt}=32.4 \;\ ft^{2}/sec.$

This means $\frac{dr}{dt}=\frac{81}{3500{\pi}}\approx{.0074}$

Now, we can find dV/dt:

$\frac{dV}{dt}=4{\pi}r^{2}\frac{dr}{dt}=4{\pi}(175) ^{2}(\frac{81}{3500\pi})=2835 \;\ ft^{3}/sec$