A water-trough is 10m long and has a cross-section which is the shape of an isosceles trapezoid

that is 30cm wide at the bottom, 80cm wide at the top, and has height 50cm. If the trough is being fi�lled with water at the rate of 0.2 m3/min, how fast is the water level rising when the water is 30cm deep?

The steps I took were as follows

b1 is the width of the water at height 0

b2 is the width of the water at height h

knowing the volume of a trapezoid is

V = 0.5(b1 + b2)hL

where L is the length of the trapezoid.

As the height increases by 1 so does the width so that means dh/dt = dw/dt

Now I tried solving for h in meters.

V = 5h(0.3 +(0.3 + h))
[0.3 + h is the width of the water at the combined height]

V = 3h + 5h^2

differentiate by d/dt

dV/dt = 3dh/dt + 10h dh/dt

(dV/dt)/(3 + 10h) = dh/dt

solve for dV/dt = 0.2 and h = 0.3

1/30 m/min = dh/dt

The answer is 1/3 cm/min but when I convert this to cm I get 10/3 cm/min.

What exactly am I doing wrong here?

Here we need to be careful while converting 0.2 m^3/min to cm ^3/min.

Please note that 0.2 m^3 =0.2x10000000= 200000cm^3

Well, it looks like you're on the right track, but there's a small math mishap in your conversion. Let's take a closer look:

You correctly find that dh/dt = 1/30 m/min. To convert this to cm/min, we'll multiply by 100, since there are 100 cm in 1 m.

So, (1/30) m/min * 100 cm/m = 100/30 cm/min.

However, we can simplify this further by dividing both the numerator and denominator by 10 to get 10/3 cm/min.

So, the correct answer is indeed 10/3 cm/min, which is approximately 3.33 cm/min.

Hope that clears things up!

It seems like you are on the right track with your approach, but there might be a mistake in your conversion from meters to centimeters. Let's go through the steps again to find the correct answer.

1. We have the equation for the volume of the trapezoidal trough as:

V = 0.5(b1 + b2)hL

2. The rate at which the trough is being filled with water is given as 0.2 m³/min.

3. We want to find how fast the water level is rising when the water is 30 cm deep, which means h = 0.3 m.

4. We need to find dh/dt, the rate of change of height with respect to time.

5. Taking the derivative of the volume equation with respect to time, we have:

dV/dt = 0.5(0 + db2/dt)hL + 0.5(b1 + b2)(dh/dt)L + 0.5(b1 + b2)h(dL/dt)

6. Since the length of the trough (L) is constant, dL/dt = 0.

7. The width of the water at height 0 (b1) is given as 0.3 m, and db1/dt = 0 since it is constant.

8. The width of the water at height h (b2) can be determined using similar triangles:

b1/h = (b1 + b2)/(h + 0.5)
0.3/0 = (0.3 + b2)/(0.3 + 0.5)

Solving for b2, we get:

b2 = 0.3(0.8)/0.3 = 0.8 m

9. Plugging in the known values into the derivative equation, we have:

0.2 = 0.5(0 + db2/dt)(0.3)(10) + 0.5(0.3 + 0.8)(dh/dt)(10)

10. Simplifying further, we get:

0.2 = (1.1)(dh/dt)

11. Finally, solving for dh/dt, we have:

dh/dt = 0.2 / 1.1 = 0.182 m/min

To convert this to centimeters per minute, we multiply by 100:

0.182 m/min * 100 = 18.2 cm/min

Therefore, the correct answer is 18.2 cm/min, not 10/3 cm/min.

Your approach to solving the problem is correct, so let's examine where the discrepancy in the final answer arises.

First, you correctly set up the volume of the trapezoid using the formula V = 0.5(b1 + b2)hL, where b1 and b2 are the widths of the trapezoid at height 0 and h, respectively, and L is the length of the trapezoid.

Then, you differentiate both sides of the equation with respect to time (t) to find how the volume changes over time, which gives you dV/dt = 3dh/dt + 10h(dh/dt).

Next, you solve for dh/dt by dividing both sides of the equation by 3 + 10h, resulting in (dV/dt)/(3 + 10h) = dh/dt.

At this point, you substitute the given values of dV/dt = 0.2 m^3/min and h = 0.3 m into the equation in order to find the value of dh/dt.

However, it seems the discrepancy arises when you convert the final answer from meters per minute (m/min) to centimeters per minute (cm/min). Let's clarify the units and do the conversion correctly:

You have found that dh/dt = (dV/dt)/(3 + 10h) = 0.2/(3 + 10(0.3)) = 0.2/(3 + 3) = 0.2/6 = 1/30 m/min.

To convert this value to centimeters per minute (cm/min), remember that 1 meter is equal to 100 centimeters. Therefore, multiply 1/30 m/min by 100 to convert it to cm/min:

(1/30 m/min) * (100 cm/m) = (1/30) * 100 cm/min = 100/30 cm/min ≈ 3.33 cm/min (rounded to two decimal places).

So, the correct conversion from meters per minute to centimeters per minute should be approximately 3.33 cm/min, not 10/3 cm/min as you previously obtained.

In conclusion, you were on the right track with your solution, but the discrepancy in the final answer was due to an incorrect conversion.

well first of all think easy way

for small changes (as dh ---> 0)
change in volume = surface area * change in height
or
dV = A dh

Now do the problem (the surface is a rectangle)
we know the length, 10 meters
what is the width when h = 30 ???
linear in depth (straight sides)
w = .30 + [(.80-.30) /.50] h
or
w = .3 + h
so for the area of the surface
A = 10(.3+h)
so
when h = .3, A = 10(.6) = 6 m^2
dV = A dh
so
dV/dt = A dh/dt
.2 m^3/min = 6 m^2 dh/dt
so
dh/dt = (.2/6) meters/min
= .03333 meters/min
= 3 1/3 cm /min