Two problems

[fbellman]

I have a Calc worksheet due soon, and I've finished most of the problems, but two of them are causing me some problems...

7. A swimmer is at a point 500m from the closest point on a straight shoreline. She needs to reach a cottage located 1800m down the shore from the closest point. If she swims at 4 m/s and she walks at 6 m/s, how far from the cottage should she come ashore so as to arrive at the cottage in the shortest time?

^ I have no idea how to even begin this problem. I'm at a complete loss.

14. An inverted conical container has a diameter of 42 in and a depth of 15 in. If water is flowing out of the vertex of the container at a rate of 35pi inches^3/sec, how fast is the depth of the water dropping when the height is 5 inches?

^ For this one, I know that whatever I do will involve the derivative. And I know that the volume of a cone is (pi*r^2*h)/3. After this, I'm stuck..

Thanks a lot for any help!

 

[galactus]

You know the volume of a cone formula.

Find \(\displaystyle \L\\\frac{dh}{dt}\) given that \(\displaystyle \L\\\frac{dV}{dt}={-35}{\pi}\ in^{3}/sec\)

Use similar triangles to get \(\displaystyle \L\\\frac{r}{h}=\frac{21}{15}\)

\(\displaystyle \L\\r=\frac{7h}{5}\).

\(\displaystyle \L\\V=\frac{{\pi}}{3}(\frac{7h}{5})^{2}h=\frac{49h^{3}{\pi}}{75}\).

You are now in terms of h(that is what you're asked to find). Differentiate with respect to time and solve for dh/dt.

 

[TchrWill]

A swimmer is at a point 500m from the closest point on a straight shoreline. She needs to reach a cottage located 1800m down the shore from the closest point. If she swims at 4 m/s and she walks at 6 m/s, how far from the cottage should she come ashore so as to arrive at the cottage in the shortest time?

Here is a similar problem process which should enable tou to solve your problem.

Crossing to a Destination in Minimum Time

A man lives in a cabin on the north edge of a 1 mile wide river. The nearest source of food and supplies is a supply depot on the other side of the river and 3 miles downstream measured parallel to the river banks.
He gets across the river in a motot boat that travels at the speed of 5 mph.
Once reaching the other side of the river, he runs the remaining distance at a speed of 10 mph.
How far downstream from his cabin (measured parallel to the river bank) should he motor in order that he reaches the depot n the minimum time?

First a picture:
Draw two horizontal lines 1 inch apart representing the river banks running east-west.
Locate point A (the cabin) toward the left end of the upper line (the north bank).
Locate point B (the depot) 3 inches to the right of point A measured parallel to the river banks.
Locate point C immediately due south of point A on the southern river bank.
Locate point D (the landing point of the boat) somewhere to the right of point C.
AD represents the path of the motor boat while crossing the river.
DB represents the running path.
Let CD = "X" and CB = L
Let the width of the river be AC = "D".
Let the speed of the motor boat be Vb.
Let the running speed be Vr.
The time to travel from A to B is given by T = (X^2 + D^2)^(1/2) + (L - X)/Vr
..........................................................................Vb
The minimum travel time can be obtained by equating the first derivitive, dT/dX to zero.
dT/dX =..........X........ ...... - 1 = 0
...........Vb(X^2 + D^2)^1/2....Vr

Then, XVr = (X^2 + D^2)^1/2
..........Vb

Squaring both sides yields X^2(Vr^2) = (X^2 + D^2)
..........................................Vb^2

Then, X^2 = X^2(Vb^2/Vr^2) + D^2(Vb^2/Vr^2) or

X^2[1 - (Vb/Vr)^2] = D^2(Vb/Vr)^2 or

X^2 = D^2(Vb/Vr)^2
........[1 - (Vb/Vr)^2]

Taking the square root of both sides yields X = .......D(Vb/Vr)......
.......................................................................[1 - (Vb/Vr)^2]^1/2

Lets apply this to our given problem.
With D = 1, L = 3, Vb = 5 and Vr = 10, we have
X = .......1(5/10)....... = .5773 miles.
......[1 - (5/10)^2]^1/2

The total time is then T = (.5773^2 + 1^2)^1/2 + (3 - .5773) = .4732 hours.
.................................................5.......................10

To reasonably verify that X = .5773 leads to the minimum time, we calculate T with X = .4773 and X = .6773.
with X = .4773, T = .4738 hr. and for X = .6773, T = .4738 hr thereby leading us to fairly safely conclude that X = .5773 miles leads to the minimum time of travel.

 

[fbellman]

Thanks to you both very much! Your suggestions helped me out a lot.

Thanks again!