A water trough is 8 m long and has a cross-section in the shape of an isosceles trapezoid that is 20cm wide at the bottom, 60 cm wide at the top, and has a height 40 cm. If the trough is filled with water at the rate of 0.2m^3/min how fast is the water level rising when the water is 10 cm deep?
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- Thread starter Annie493
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Annie493 said:A water trough is 8 m long and has a cross-section in the shape of an isosceles trapezoid that is 20cm wide at the bottom, 60 cm wide at the top, and has a height 40 cm. If the trough is filled with water at the rate of 0.2m^3/min how fast is the water level rising when the water is 10 cm deep?
First draw a sketch.
Find the width of the trapezoid at a height h.
Then find the volume of the element at that height.
Please show us your work, indicating exactly where you are stuck - so that we may know where to begin to help you
Annie493 said:A water trough is 8 m long and has a cross-section in the shape of an isosceles trapezoid that is 20cm wide at the bottom, 60 cm wide at the top, and has a height 40 cm. If the trough is filled with water at the rate of 0.2m^3/min how fast is the water level rising when the water is 10 cm deep?
Again, draw the trapezoid and label the 'givens'. Use a proportion to relate 's' to the height of the water.
\(\displaystyle \frac{s}{h}=\frac{20}{20}=1\). Please pardon my diagram. It is surely not to scale, but relays the idea.
\(\displaystyle s=h\). This can be seen from the 20-40-60 dimensions.
Width of water at depth h: \(\displaystyle w=20+2s\)
\(\displaystyle \frac{dV}{dt}=\frac{1}{5} \;\ m^3/min\)
Now, express w in terms of h. Use the formula for the area of a trapezoid and express it in terms of h. Do not forget the length of the trough. differentiate w.r.t time and find dh/dt.