CALC question.. need help please!?

Question

A trough is 5 feet long, and its vertical cross sections are inverted isosceles triangles with base 2 feet and height 3 feet. Water is being siphoned out of the trough at the rate of 2 cubic feet per minute. At any time t, let h be the depth and V be the volume of water in the trough.

(a) Find the volume of water in the trough when it is full.

(b) What is the rate of change in h at the instant when the trough is 1/4 full by volume?

(c) What is the rate of change in the area of the surface of the water at the instant when the trough is 1/4 full by volume?

Answer 1

Part of this problem involves getting a good picture in your mind of the object. In this case, it is not too difficult.

a) This is just a simple area problem, no calculus needed. You know the volume of a cylinder is pi*r^2*h. H is just the length of the cylinder and pi*r^2 is the area of a cross section. Applying this knowledge to our trough, we just apply the area of a triange formula for the cross sections and multiply by the length:
V = (1/2)(2)(3)(5) = 15 ft^3.

b) This is clearly a rate of change problem, but it is not as straightforward. We need to use the volume formula derived above and then take a derivative with respect to time.

V=(1/2)B*H*L, where B=base of the triangle cross sections, H=height, and L=length. Now we differentiate with respect to time. V, b, and h are going to vary with time, while l remains constant, so the derivative is:
dV/dt = (1/2)L[B*dH/dt + H*dB/dt]
Now, we have do some geometry of Isosceles triangles in order to find relationships between the base and height, as well as between the rate of changes of the base and height. The reason is because we are only given the rate of change of Volume, so we need to elliminate some of the variables.

However, there is really a better way to deal with this. I went through the above to clarify the technical process involved. What we should really do before we differentiate the volume formula is subsitute the height for a term involving the base. The reason we can do this is because for an isosceles triangle, the height = (3/2) * Base or Base = 2/3 * height. What is meant here is that when the water is a certain heighth, the base of the cross section will be 2/3 that height. As a check, when the water is 3 ft. up, the base of the water is 2 ft, which was given in the problem. We will make this substitution in the volume formula:

V = (1/2) H * 2/3*H * L = 1/3 * H^2 * L. Then we differentiate:
dV/dt = 2/3 * L * H * dH/dt. We are trying to solve for dH/dt, so:

dH/dt = (dV/dt)/(2/3 * H * L).

First we need to find H. H again will require use of a little geometry. We already figured out a full volume of water as 15, so 1/4 of the volume is 15/4 = 3.75. Then solve for H:
V = (1/2) B * H * L = 1/3 * H^2 * L, which was derived above.
H = root(V/[(1/3) * L]) = root(3.75/[(1/3) * 5]) = 1.5

The other variables are:
dV/dt = -2 this was given is the problem as the rate at which the water was being removed.
L = 5 ft.
Now we can plug them into the formula to find:
dH/dt = (dV/dt)/(2/3 * H * L) = (-2)/(2/3 * 1.5 * 5) = -.4 ft/min

In this problem, the writers were nice enough to use consistant units (feet for length and minutes for time). You just have to make sure you follow the same convention in you answer. Notice that rate of change of height is going to be in units of length of units of time. Since throughout the problem we used feet for length and minutes for time, our answer is going to be in units of feet/minute. If the author used something else, say inches for the length of the trough, we would need to convert to feet to ensure consistant units. Notice also that the rate of change of height is negative, which we should expect considering that the water is being taken out of the trough.

c) This will involve similar steps as above, just different units. The best way to get this is to find some expression that relates the volume with the area of the surface of the water. The surface of the water is just a rectangle, with length equal to the trough and a width equal to the Base of the 'cross section' triange we have been using all along. Therefore, the area of the surface of the water is:

A = B * L, using the same notation for B and L as above.

The Volume formula again is:
V = (1/2) B * H * L = (1/2) B * (3/2)B * L

This time you will find the rate of change of the Base, just like we did for the height, using the Volume formula. You will then differentiate the Area formula with respect to time (remember L is constant) and you should have dB/dt to plug-in in order to find dA/dt.

Answer 2

The question I iced over up on in the final Calc III classification very final examination I took grew to become into one the place you had to alter from Cartesian coordinates to cylindrical coordinates, or perhaps it grew to become into around coordinates to cylindrical coordinates. I had in specific studied that distinctive concern because of the fact I KNEW that i might in all probability freeze up when I had to alter coordinate structures, and that i did. I took this classification for my B.S. in actual geology, when I had taken Engineering math I and II (Calc IV and V) as standards for an M.S. in geophysics. decelerate and take it sluggish, I in all probability nevertheless have math stress issues, and my eyes sense like they're glazing over if there are 3 or extra integrals on a mag internet site, yet now seem at necessary equations as an relaxing challenge/puzzle to untangle, no longer a chore. I took Calc I, II, and III 2 circumstances at 3 diverse universities, yet I have been given an A in my first Calc IV classification at Tulane. i attempted to take linear algebra three times (Calc III prerequisite), and withdrew each and every time. make an effort to earnings and understand. then you particularly could no longer could repeat classes. i'm no longer stupid in math, whether it took an somewhat long term for me to realize that. A 620 on the maths area of the familiar flair GRE grew to become into what ultimately confident that i'm no longer stupid in math, I in basic terms want extra time to particularly understand the way it works than numerous human beings.

Answer 3

Let
V = volume of water in trough
b = base
h = height water
L = length trough
S = surface area

Given

dV/dt = -2 ft³/min

(a) Find the volume of water in the trough when it is full.

V = (1/2)bhL = (1/2)*3*2*5 = 15 ft³

(b) What is the rate of change in h at the instant when the trough is 1/4 full by volume?

Find dh/dt.

b/h = 2/3
b = 2h/3

V = (1/2)(2h/3)h*5 = 5h²/3
dV/dh = 10h/3

dh/dt = (dV/dt) / (dV/dh) = -2/(10h/3) = -3/(5h)

When the trough is 1/4 full by volume
h = (1/2)(3) = 3/2

dh/dt = -3/(5h) = -3/[5*(3/2)] = -2/5 ft/min

(c) What is the rate of change in the area of the surface of the water at the instant when the trough is 1/4 full by volume?

V = (1/2)bhL = (5/2)bh
h/b = 3/2
h = 3b/2
V = (5/2)bh = (5/2)b(3b/2) = 15b²/4
b² = 4V/15
b = 2√(V/15)

S = bL = 5b = 5*2√(V/15) = 10√(V/15)
S = 10(√15V)/15 = (2/3)(√15V)

dS/dV = (1/2)(2/3)√(15/V) = (1/3)√(15V)

dS/dt = (dS/dV)(dV/dt) = (1/3)√(15V)(-2)
dS/dt = (-2/3)√(15V)

Now plug in V/4 = 15/4

dS/dt = (-2/3)√[15(15/4)] = (-2/3)√[225/4)]
dS/dt = (-2/3)(15/2) = -5 ft²/min

Answer 4

answer = square root of ( i have no idea, sorry )