Please help me!!

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Apr 9th 2006, 08:37 PM
LilDragonfly

Please help me!!

Hi everyone. I just have three questions that I can't answer and if anyone could spare the time to help me I will be very thankful :)

1.
A pile of sand is in a roughly conical pile. The slant edge of the pile is 6.0m.

a) what is the height of the pile when the volume of sand in it is at a maximum?

b) a truck operator contracted to shift the sand wants to know how many cubic metres there are. What answer should he be given? Justify your answer with regerence to the situation described.

2.
Nico's bubble blowing machine blows spherical bubbles whose surface area, S, increases at the rate of 23 cm²/s.
What is the rate of increase of the radius, r, of the bubble when the radius is 5 cm?

3.
The height in metres of a bullet fired up into the air after t seconds is given by h = 2000 t - 200 t²
Find the maximum height the bullet reaches. (A test to find the maximum height is required)
Apr 10th 2006, 08:03 AM
ThePerfectHacker

Quote:

Originally Posted by LilDragonfly

Hi everyone. I just have three questions that I can't answer and if anyone could spare the time to help me I will be very thankful :)

1.
A pile of sand is in a roughly conical pile. The slant edge of the pile is 6.0m.

a) what is the height of the pile when the volume of sand in it is at a maximum?

b) a truck operator contracted to shift the sand wants to know how many cubic metres there are. What answer should he be given? Justify your answer with regerence to the situation described.

The formula for volume of a cone is,
$\frac{1}{3}\pi r^2 h$ .
Since, $r^2+h^2=6^2=36$ (Pythagorean Theorem),
we can express volume in terms of height,
$V=\frac{1}{3}\pi (36-h^2)h=\frac{1}{3}\pi (36h-h^3)$
Taking its derivative,
$V'=\frac{1}{3}\pi (36-3h^2)$ and finding critical points,
$\frac{1}{3}\pi (36-3h^2)=0$ thus,
$h^2=12$ thus,
$h=\pm \sqrt{12}$
reject negative, thus, $h=\sqrt{12}$ .
To prove that this is a maximum we use the second derivative test, first we need the second derivative,
$V''=\frac{1}{3}\pi(-6h)$
We can see that at $h=\sqrt{12}$ it is negative thus we have a maximum.
--------
Next, you should tell the truck operator the maximum number because then he would definetly remove all the debris. Which is the Volume evaluated at this height.
Since, $V=\frac{1}{3}\pi (36h-h^3)$
at $h=\sqrt{12}$
Thus,
$\frac{1}{3}\pi (36\sqrt{12}-(\sqrt{12})^3)$
Thus,
$\frac{1}{3}\pi (36\sqrt{12}-12\sqrt{12})$
Thus,
$\frac{1}{3}\pi (24\sqrt{12})$
Thus,
$8\pi\sqrt{12} \mbox{m}^3$
Of course, you cannot tell the truck operator this number he would have no idea what you mean.
$\approx 87 \mbox{m}^3$
Apr 10th 2006, 08:09 AM
ThePerfectHacker

Quote:

Originally Posted by LilDragonfly

2.
Nico's bubble blowing machine blows spherical bubbles whose surface area, S, increases at the rate of 23 cm²/s.
What is the rate of increase of the radius, r, of the bubble when the radius is 5 cm?

You know that,
$S=4\pi r^2$
Thus, taking derivative with respect to time,
$\frac{dS}{dt}=8\pi r\frac{dr}{dt}$
Since the problem says,
$\frac{dS}{dt}=23$ and, $r=5$
you have,
$23=40\pi \frac{dr}{dt}$
Thus, solving,
$\frac{dr}{dt}\approx .18 \frac{\mbox{m}}{\mbox{sec}}$
Apr 10th 2006, 08:13 AM
ThePerfectHacker

Quote:

Originally Posted by LilDragonfly

3.
The height in metres of a bullet fired up into the air after t seconds is given by h = 2000 t - 200 t²
Find the maximum height the bullet reaches. (A test to find the maximum height is required)

You need the find the critical points. Set derivative to zero.
$2000-400t=0$ thus, t=5[/tex]
Next, take the derivative of this (second derivative),
$h''=-400$ , notice it is negative. Thus, the point is maximum. Meaning, that $t=5$ maximum height occurs. But the problem asks not for the time rather for the height the bullet reaches which is the function evaluated at $t=5$ ,
$h(5)=2000(5)-200(5)^2=5000\mbox{m}$
Apr 10th 2006, 02:23 PM
LilDragonfly

Thank you for your time and effort explaining this to me :)