A trough is 14 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1 ft. If the trough is filled with water at a rate of 13 ft3/min, how fast (in ft/min) is the water level rising when the water is 5 inches deep?

ratio of width to depth of water = 4/1 = 4

if depth is 5 in = 5/12 ft then w = 4 * (5/12) = 5/3 ft
area of surface = w L = (5/3)*14 = 23.33 ft^2
dh * w L = d volume
dh / dt * wL = d volume / dt = 13 fT^3/min
dh/dt = 13 ft^3/min / 23.33ft^2 = 0.557 ft/min

Well, let's dive into this problem, shall we?

First, we need to convert the height of 5 inches into feet. Since there are 12 inches in a foot, 5 inches would be 5/12 feet.

Now, we know the volume rate at which water is filling the trough is 13 ft3/min.

Since the trough has the shape of two isosceles triangles, we can find the area of one of the triangles by using the formula for the area of a triangle: (base * height) / 2. The base of the triangle is 4 feet and the height is 1 foot.

So, the area of one triangle is (4 * 1) / 2 = 2 square feet.

Since there are two triangles, the total cross-sectional area of the ends of the trough is 2 square feet * 2 = 4 square feet.

To find how fast the water level is rising, we need to find the rate at which water volume is increasing. This can be done by multiplying the cross-sectional area of the trough ends by the rate at which water is filling the trough.

So, the rate at which water volume is increasing is 4 square feet * 13 ft3/min = 52 ft3/min.

Now, we can divide the rate at which water volume is increasing by the cross-sectional area of the trough ends to find how fast the water level is rising.

So, the water level is rising at a rate of 52 ft3/min / 4 square feet = 13 feet per minute when the water is 5 inches deep.

I hope that didn't make you too "trough" to handle!

To find how fast the water level is rising, we need to determine the rate at which the volume of the water is increasing with respect to time. This can be found using related rates.

Let's start by drawing a diagram of the trough:

_________
/ \
/| |\
/ | | \
/ | | \
/ | | \
/ | | \
/___|_________|____\

Notice that if we cut the triangle-shaped ends off, the remaining shape is a rectangle. Let's create a rectangle with the same dimensions as the trough:

_________
| |
| |
| |
| |
| |
|_________|

The length of this rectangle is still 14 ft, and its width is now 4 ft.

The volume of water in the trough can be found by multiplying the area of the rectangle by the height of the water. Let's denote the height of the water as "h".

Volume = Area * Height
V = (Length * Width) * h
V = (14 ft * 4 ft) * h = 56h ft³

We are given that water is flowing into the trough at a rate of 13 ft³/min. This means that the rate at which the volume of water is increasing with respect to time is 13 ft³/min.

We are asked to find how fast the water level is rising when the water is 5 inches deep. Since the height of the water is in inches, we need to convert it to feet:

1 ft = 12 inches
5 inches = (5/12) ft

Now we have the value of h (height) and we want to find the rate at which h is changing with respect to time (dh/dt).

To do this, we differentiate both sides of the equation V = 56h with respect to time (t):

dV/dt = d/dt (56h)
13 ft³/min = 56 (dh/dt)

Now we can solve for dh/dt, which represents the rate at which the water level is rising:

dh/dt = 13 ft³/min / 56
dh/dt ≈ 0.232 ft/min

Therefore, the water level is rising at a rate of approximately 0.232 ft/min when the water is 5 inches deep.

Did you make a sketch?

I made a sketch of an isosceles triangle with width of 4 ft and a height of 1 ft

At some time of t min, let the width of the water level be 2r and h
then for the water in the trough at that moment,
V = (1/2)(2r)(h)(14) = 14 hr

we have similar triangles: so r/h = 2/1 ===> r = 2h
V = 14(2h)(h) = 28 h^2
dV/dt = 56 h dh/dt
so for my given data ...
13 = 56(5/12)dh/dt
dh/dt = 156/280 ft/min = 39/70 ft/min
= appr .557 ft/min

better check my arithmetic

LOL Your arithmetic checks :)