A trough is 10 ft long and its ends have the shape of isosceles triang

Ernest Ryland 2021-12-18 Answered
A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 12ft3min, how fast is the water level rising when the water is 6 inches deep?
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Expert Answer

Natalie Yamamoto
Answered 2021-12-19 Author has 22 answers
Step 1
When the water is x feet deep, let us assume that the isosceles cross section formed by the water is b feet wide
Using the property of similar triangles, we can write
b3=x1
Multiply both sides by 3
b=3x
The area of the cross section formed by the water is
A=12(base)(height)
Substitute 3x in place for base and x for height
A=3x22
Volume of the water=(length of the trough)×(Area of the cross section)
V(x)=10×3x22=15x2
Step 2
To find the rate at which the water level is rising, differentiate with respect to time
dVdt=d(15x2)dt
Use the chain rule in the right-hand side
dVdt=d(15x2)dx×dxdt
Use the power rule to differentiate 15x2
dVdt=15×2x21×dxdt=30x×dxdt
Substitute x=0.5 because 6inches=0.5feet and dVdt=12
12=30×0.5×dxdt
12=15×dxdt
dxdt=1215=45=0.8ftmin
The water level is rising at the rate of 0.8ftmin
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veiga34
Answered 2021-12-20 Author has 32 answers

Step 1
Given Information:
V=b×h×l2
dVdt=12ft3min
when h=6 inchsh=0.5ft
dhdt=?
31=bhb=3h
v=b×h×102v3h×h×102V=15h2
Step 2
dVdt=dVdh×dhdt
Sub in for h=0.5ft and dVdt=12ft3min
dhdt=45=0.8ftmin

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nick1337
Answered 2021-12-27 Author has 575 answers

Step 1
Let V (t) =he volume of water in the trough at time t.
We knowdVdt=12ft3minsince the amount of water in the trough is increasing (“filled”) at the rate of12ft3min
Let h=the “depth” of water in the trough at time t. Note that this depth is in fact the height of the triangle that the water makes on the end of the trough.
We want dhdtwhen h=12ft
We need an equation relating V and h.
V=(area of base of trough)×(height of trough)
=(area of theof water)×(10 ft)=(12×base of theof water×height of theof water)×(10 ft)
To find the base of the of water in terms of h, we use the similar triangles shown at right.
=(12×3 h×h)(10)=15h2
We differentiate with respect to t
ddt[V]=ddt[15h2]

dVdt=30h×dhdt
substituting
12=30×12dhdt
1215=dhdt
or
45=dhdt
Thus the water level is rising at the rate of 45ftminwhen the water is 6 inches deep.

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