A water trough on a farm has an isosceles triangle croos section which is 60 cm across the top and 20m deep.the trough is 3m long . If itbis empty and then filled at a rate of 0.5 m3/min , how fast does the water level rise when the deepest point is 9cm ?
This problem is similar to one of your previous questions:
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To find how fast the water level rises when the deepest point is 9 cm, we need to determine the rate of change of the water level with respect to time. Let's break down the problem into smaller parts:
Step 1: Find the volume of water in the trough when the water level is at a certain height.
Step 2: Determine how the volume of water changes with respect to time.
Step 3: Use the provided rate of filling the trough to calculate the rate of change of height.
Step 1: Find the volume of water in the trough when the water level is at a certain height.
The cross-section of the trough is an isosceles triangle. Since the base of the triangle is 60 cm and the height is 20 m, we can consider the triangle to be a right triangle. The height of the triangle is the same as the depth of the trough, which is 20 m.
The formula for the area of a right triangle is: Area = (1/2) * base * height
In this case, the base is 60 cm (or 0.6 m), and the height is 20 m (since the trough is 20 m deep).
Therefore, the area of the triangle is: Area = (1/2) * 0.6 * 20 = 6 m^2
Step 2: Determine how the volume of water changes with respect to time.
The volume of water in the trough can be calculated by multiplying the cross-sectional area by the width (length) of the trough. In this case, the width is 3 m.
So, the volume of water in the trough is: Volume = area * width = 6 * 3 = 18 m^3
Step 3: Use the provided rate of filling the trough to calculate the rate of change of height.
The trough is being filled at a rate of 0.5 m^3/min. To find the rate of change of height, we need to calculate how fast the volume is changing with respect to time when the volume is 18 m^3. This can be done using the chain rule of differentiation.
Let h represent the height of the water in the trough (in meters), and let t represent the time (in minutes).
From Step 1, we know that the area of the triangle remains constant at 6 m^2.
From Step 2, we know that the volume of water V is given by V = 6h.
Differentiating both sides of the equation with respect to time t yields:
dV/dt = d(6h)/dt
= 6 * dh/dt
Given that dV/dt is 0.5 m^3/min, we can substitute the values:
0.5 = 6 * dh/dt
To find dh/dt, divide both sides of the equation by 6:
dh/dt = 0.5 / 6
= 1/12 m/min
Therefore, the water level rises at a rate of 1/12 meters per minute when the deepest point is 9 cm.