Gasoline is pumped from the tank of a tanker truck at a rate of 20L/s. If the tank is a cylinder 2.5m in diameter and 15m long, at what rate is the level of gasoline falling when the gasoline in the tank is 0.5m deep?

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Gasoline is pumped from the tank of a tanker truck at a rate of 20L/s. If the tank is a cylinder 2.5m in diameter and 15m long, at what rate is the level of gasoline falling when the gasoline in the tank is 0.5m deep?

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Answer 1 · 2018-03-27T16:17:23

The level is falling at a rate of $0.8627 m m {sec}^{- 1}$ $0.8627 mm sec^-1$

Explanation:

Need to consider the volume of the fluid in the tanker, making a sketch of sector which is formed within the tank[ Which I am unable to supply] It can be seen that a triangle is formed from the centre of the tanker to the level of the fluid and where the level of the fluid touches the sides of the tank.

We need to find the volume of this sector,which is given by multiplying the area of this sector by it's length.

Let the height of the fluid= $h$ $h$ ,we can see that the height $h$ $h$ of the fluid can be expressed in terms of the radius, If the level of the fluid is $h$ $h$ and we know that the radius of the of the tank = $1.25$ $1.25$ then we may say that, $[r - h = 0.75]$ $[r-h=0.75]$ [ since $h = 0.5]$ $h =0.5]$ i.e, $[r + h = 0.75]$ $[r+h=0.75]$

The area of a sector = $\frac{r^{2} θ}{2}$ $[r^2theta]/2$ , where $θ$ $theta$ is in radians. We can see that a horizontal line through the centre of the tank, divided by two, is one leg of a right triangle of length $r = 1.25$ $r=1.25$

The other leg of the right triangle is formed by a line vertically through the centre of the tank with a length of $0.75$ $0.75$ [To the surface of the fluid]

So it can be said that $cos θ = \frac{0.75}{1.25}$ $Costheta=0.75/1.25$ , ${cos}^{- 1} θ = 0.9273 r a d i a n s$ $cos^-1theta= 0.9273 radians$ , But we need to multiply this by two to give the total angle of the sector, = $1.8546 r a d i a n s$ $1.8546 radians$

Volume in tank will then = $\frac{r^{2} θ}{2}$ $[r^2theta]/2$ , = $\frac{1}{2} [{[h + 0.75]}^{2}$ $1/2[[h+0.75]^2$ [10], since $r = [h + .75]$ $r=[h+.75]$ and the length of the tanker is $10$ $10$

Therefore volume= $5 {[h + .75]}^{2} [1.8546]$ $5[h+.75]^2[1.8546]$ , now need to differentiate this to find rate of change of height and volume with respect to time.

$d \frac{v}{d t} = 10 [1.8546] [h + .75] d \frac{h}{d t}$ $dv/dt=10[1.8546][h+.75]dh/dt$ , but we know that $d \frac{v}{d t} = 20$ $dv/dt =20$ litres a second, however we need the height in terms of metres [or millimetres], since there are $1000$ $1000$ litres in a cubic meter, $d \frac{v}{d t} = \frac{20}{1000}$ $dv/dt=20/1000$ [cubic meters sec^-1]

Therefore, $\frac{1}{50}$ $1/50$ = $10 [1.8546] [h + .75] d \frac{h}{d t}$ $10[1.8546][h+.75]dh/dt$ , and solving for $d \frac{h}{d t}$ $dh/dt$

$d \frac{h}{d t}$ $dh/dt$ = $\frac{1}{[50] [10] [1.8546] [h + .75]}$ $1/[[50][10][1.8546][h+.75]]$ and when $h = 0.5$ $h=0.5$ this expression evaluated will give the rate of change in meters per ${sec}^{_} 1$ $sec^_1$ . I have just converted the answer to mm per ${sec}^{- 1}$ $sec^-1$ Apologies for length of answer, It's the only way I know to express it.

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Gasoline is pumped from the tank of a tanker truck at a rate of 20L/s. If the tank is a cylinder 2.5m in diameter and 15m long, at what rate is the level of gasoline falling when the gasoline in the tank is 0.5m deep?

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