A horizontal trough is 40 cm long and its ends are in the form of an isosceles trapezoids with an altitude of 20 cm, a lower base of 20 cm and an upper base of 3 cm. Water is being poured into the trough at the rate of 5 cm/sec. How fast is the water level rising when the water is 10 cm deep?
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To find how fast the water level is rising, we need to calculate the derivative of the water level with respect to time. This can be done using related rates.
Let's denote the water level as h (in cm) and the time as t (in seconds). We are given that water is being poured into the trough at a rate of 5 cm/sec, so dh/dt = 5 cm/sec.
Now, our goal is to find dh/dt when the water level is 10 cm deep, so h = 10 cm.
First, let's find the expression for the area of water in the trough. The trough can be divided into two parts: a rectangle and an isosceles trapezoid.
The rectangle has a length of 40 cm and a width of h cm.
The area of the rectangle is A_rect = length * width = 40 * h = 40h cm^2.
The isosceles trapezoid has an upper base of 3 cm, a lower base of 20 cm, and an altitude of 20 cm. We can find the length of the trapezoid (l) using similar triangles.
By similarity, we can set up the following proportion:
l / (l - 20) = 20 / h
Cross-multiplying, we get:
l * h = 20 * (l - 20)
lh = 20l - 400
lh - 20l = -400
l(h - 20) = -400
l = -400 / (h - 20)
Since length cannot be negative, we take the absolute value:
l = 400 / (20 - h)
Now, the area of the isosceles trapezoid can be calculated as:
A_trapezoid = (1/2) * (upper base + lower base) * altitude
= (1/2) * (3 + 20) * 20
= 23 * 20
= 460 cm^2
The total area of water in the trough is the sum of the areas of the rectangle and the trapezoid:
A = A_rect + A_trapezoid
= 40h + 460
Now, we differentiate both sides of the equation with respect to time:
dA/dt = d(40h + 460)/dt
The left side represents the rate at which the area is changing, which is dA/dt. And the right side represents the rate at which the water level is changing, which is dh/dt. Since the area of the trough is changing only due to the changing water level, the rate at which the area changes is equal to the rate at which the water level changes multiplied by the width of the trough (40 cm).
dA/dt = 40 * dh/dt
Now, let's substitute the values we have:
40 * dh/dt = d(40h + 460)/dt
Differentiating the right side of the equation:
40 * dh/dt = 40 * dh/dt + d(460)/dt
The derivative of a constant (460) is zero, so the equation simplifies to:
40 * dh/dt = 40 * dh/dt
Now, we can cancel out the common terms on both sides:
dh/dt = dh/dt
This shows that the rate at which the water level is changing (dh/dt) is equal to the rate at which the water level is changing (dh/dt).
In other words, the rate at which the water level is rising is 5 cm/sec, regardless of the depth of the water.
Therefore, when the water is 10 cm deep, the water level is rising at a rate of 5 cm/sec.