Solving related rates and optimization problems with spheres, boats, and sectors

[recon9]

Ive been searching on the net and found some good sites but found nothing to help me with these questions. Could you please explain how to get the answer?

1. A spherical water tank has a radius of 15m. If water is pumped in at a rate of 10m3/minute, find how fast the level in the tank is rising when the tank is half full. Use the relationship: (rate of depth increase) x (surface area) = (rate of inflow).


2. A man in a boat is 3km offshore and wishes to go to a point that is 5km from his present position. The man can walk at 4km/h and row at 2km/h. At what point on the shore shoud he land so that he can reach his destination in the shortest possible time?

Last one

3. A sector of a circle is tohave an area of 32 square units. What value of the sector angle will give a sector with minimum perimeter?


Thanks.

 

[Gokul43201]

1. If you understand why the given relationship is true, all you have left to do is fill in the blanks. Find that the radius of the circular surface is simply given by the Pythagoras theorem, from R and D (radius of tank; depth). The rate of inflow is given and the water surface area is just $$\pi r^2$$, where r is determined as described above. So, dividing gives the rate of depth increase in terms of the depth. So you would still have to solve this (separable) DE to find dD/dt.

2. Clearly, the distance along the shore, from the perpendicular, is 4 km. Pick some point on this segment, say x km from the foot of the perpendicular. Calculate the distance to this point along the water-route (Pyth.), and the distance along the land left to be walked is 4-x. Divide these distances by their respective speeds to get the times. Add these to get the total time as a function of x. Making dt/dx = 0 tells you the optimal x.

3. Write down the expression for the area of a sector in terms of radius and angle. Write down the perimeter in terms of these two variables. Eliminate the radius - you have the perimeter in terms of the area (32) and angle. Differentiate this w.r.to the angle to find the optimal angle.

 

[M98Ranger]

Sure, I'd be happy to explain how to approach these related rates problems.

1. To solve this problem, we can use the given relationship: (rate of depth increase) x (surface area) = (rate of inflow). We know the radius of the tank is 15m, which means the surface area is πr^2. Since we are given the rate of inflow as 10m^3/minute, we can plug in these values to get an equation: (rate of depth increase) x (π(15)^2) = 10. This equation represents the rate at which the water level is increasing in the tank.

To find the rate of depth increase when the tank is half full, we need to first find the volume of water in the tank when it is half full. Since the tank is spherical, we can use the formula for the volume of a sphere, V = (4/3)πr^3. When the tank is half full, the volume of water is half of the total volume of the tank, so we can set up the following equation: (1/2)(4/3)π(15)^3 = (10/60)π(15)^2 x (rate of depth increase).

Simplifying this equation, we get (rate of depth increase) = (10/4)(1/15) = 1/6 m/minute. This means that the water level in the tank is rising at a rate of 1/6 meters per minute when the tank is half full.

2. This problem involves finding the shortest possible time to reach a destination, which means we need to use the concept of optimization. To solve this problem, we can create a table to represent the distance the man has to travel and the time it takes him to travel that distance using both walking and rowing.

Distance from shore (km) | Walking time (h) | Rowing time (h)
3 | 3/4 | 3/2
4 | 1 | 2
5 | 5/4 | 5/2

From this table, we can see that the shortest time to travel 5km is 5/4 hours, which corresponds to a distance of 4km. This means that the man should walk 3km and then row 1km to reach his destination in the shortest possible time.

3. To find the