Test Review Questions

[thenextangusyoung]

we have 5 questions, and I can do 3 no problem. I'm stuck on the last two.

1.) Driver A had been leading archrival driver B for a while by a steady 3 miles. Only 2 miles from the finish, A ran out of gas decelerated thereafter at a rate proportional to the square of his remaining speed. 1 mile later, A's speed was exactly halved. If driver B's speed remained constant, who won the race?

2.) 1) Assuming the efflux of water (volume per unit time) through an orifice in the bottom of a tank is
proportional to the product of the area of the orifice and the square root of the depth of the water, the
differential equation is :

A*dh/dt = - k*B*h^(1/2)

square root of h----^

where h (ft) is the depth of the water and A (ftxft) is the area of water surface at any time t (sec), and B
(ftxft) is the area of the orifice. The constant k of proportionality may be determined empirically.
a) Find the time required to empty a cubical tank whose edge is 4 feet. The tank has a hole 2
inches in diameter in the bottom and the tank is originally full of water. Let k=4.8
b) A funnel has the shape of a right circular cone with vertex down and is full of water. If half the
volume of water runs out in time t, find the time required to empty.

 

[galactus]

2.) 1) Assuming the efflux of water (volume per unit time) through an orifice in the bottom of a tank is
proportional to the product of the area of the orifice and the square root of the depth of the water, the
differential equation is :

\(\displaystyle A\frac{dh}{dt} = - k\cdot B\cdot \sqrt{h}\)


where h (ft) is the depth of the water and A (ftxft) is the area of water surface at any time t (sec), and B
(ftxft) is the area of the orifice. The constant k of proportionality may be determined empirically.
a) Find the time required to empty a cubical tank whose edge is 4 feet. The tank has a hole 2
inches in diameter in the bottom and the tank is originally full of water. Let k=4.8

The orifice is measured in inches instead of feet. So, to keep consistent units, the orifice has area \(\displaystyle B={\pi}(\frac{1}{12})^{2}=\frac{\pi}{144} \;\ ft^{2}\)

The area of the water surface of the cubical tank is \(\displaystyle A=16 \;\ ft^{2}\)

So, the DE becomes:

\(\displaystyle 16\frac{dh}{dt}=\frac{-24}{5}\cdot \frac{\pi}{144}\sqrt{h}\)

Integration leads to:

\(\displaystyle h=\left(C-\frac{\pi t}{960}\right)^{2}\)

Using the fact that at time t=0, the water is h=4, then we find C=2.

\(\displaystyle h=\left(2-\frac{\pi t}{960}\right)^{2}\)

Now, to find how long to empty the tank, set h=0 and solve for t.

I notice gravity is not considered here, so I assume it has been incorporated into k.

Another form commonly used for draining a tank is \(\displaystyle A_{w}\frac{dh}{dt}=-kB\sqrt{2gh}\).

See how the gravitational constant is implemented here. I suppose in this problem, the \(\displaystyle \sqrt{2g}=\sqrt{64}=8\) has been used in k.

For the cone problem, I see no information. Is it just a general case using a cone with radius R and height H?. Is k the same as with the cube?. Either way, use similar triangles as was done with related rates back in calc I.