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

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 $12\frac{f{t}^{3}}{min}$, how fast is the water level rising when the water is 6 inches deep?
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Natalie Yamamoto
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
$\frac{b}{3}=\frac{x}{1}$
Multiply both sides by 3
$b=3x$
The area of the cross section formed by the water is
$A=\frac{1}{2}\text{(base)(height)}$
Substitute 3x in place for base and x for height
$A=\frac{3{x}^{2}}{2}$
$\text{Volume of the water}=\text{(length of the trough)}×\text{(Area of the cross section)}$
$V\left(x\right)=10×\frac{3{x}^{2}}{2}=15{x}^{2}$
Step 2
To find the rate at which the water level is rising, differentiate with respect to time
$\frac{dV}{dt}=\frac{d\left(15{x}^{2}\right)}{dt}$
Use the chain rule in the right-hand side
$\frac{dV}{dt}=\frac{d\left(15{x}^{2}\right)}{dx}×\frac{dx}{dt}$
Use the power rule to differentiate $15{x}^{2}$
$\frac{dV}{dt}=15×2{x}^{2-1}×\frac{dx}{dt}=30x×\frac{dx}{dt}$
Substitute $x=0.5$ because $6\text{inches}=0.5\text{feet}$ and $\frac{dV}{dt}=12$
$12=30×0.5×\frac{dx}{dt}$
$12=15×\frac{dx}{dt}$
$\frac{dx}{dt}=\frac{12}{15}=\frac{4}{5}=0.8\frac{ft}{min}$
The water level is rising at the rate of $0.8\frac{ft}{min}$
###### Not exactly what you’re looking for?
veiga34

Step 1
Given Information:
$V=\frac{b×h×l}{2}$
$\frac{dV}{dt}=12\frac{f{t}^{3}}{min}$
when
$\frac{dh}{dt}=?$
$\frac{3}{1}=\frac{b}{h}⇒b=3h$
$v=\frac{b×h×10}{2}⇒v\frac{3h×h×10}{2}⇒V=15{h}^{2}$
Step 2
$\frac{dV}{dt}=\frac{dV}{dh}×\frac{dh}{dt}$
Sub in for $h=0.5ft$ and $\frac{dV}{dt}=12\frac{f{t}^{3}}{min}$
$\frac{dh}{dt}=\frac{4}{5}=0.8\frac{ft}{min}$

###### Not exactly what you’re looking for?
nick1337

Step 1
Lethe volume of water in the trough at time t.
We know$\frac{dV}{dt}=12\frac{f{t}^{3}}{min}$since the amount of water in the trough is increasing (“filled”) at the rate of$12\frac{f{t}^{3}}{min}$
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 $\frac{dh}{dt}$when $h=\frac{1}{2}ft$
We need an equation relating V and h.
$V=\text{(area of base of trough)}×\text{(height of trough)}$
$=\text{(area of the}∆\text{of water)}×\text{(10 ft)}$
To find the base of the $∆$of water in terms of h, we use the similar triangles shown at right.
$=15{h}^{2}$
We differentiate with respect to t
$\frac{d}{dt}\left[V\right]=\frac{d}{dt}\left[15{h}^{2}\right]$

$\frac{dV}{dt}=30h×\frac{dh}{dt}$
substituting
$12=30×\frac{1}{2}\frac{dh}{dt}$
$\frac{12}{15}=\frac{dh}{dt}$
or
$\frac{4}{5}=\frac{dh}{dt}$
Thus the water level is rising at the rate of $\frac{4}{5}\frac{ft}{min}$when the water is 6 inches deep.