# Answer on Calculus Question for Abhilash

Question #2103

A water trough is 5 m long and has a cross-section in the shape of an isosceles trapezoid that is 40 cm wide at the bottom, 100 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.1 m3/min how fast is the water level rising when the water is 20 cm deep?

_____________ m/min

_____________ m/min

Expert's answer

Let's find the angle between the top and the side wall.

Denote the top side as a (a= 100 cm), the bottom side as b (b =40 cm),

the height of trapesoid as h0 (h0=60 cm), the height of the water as h (h = 20 cm) :

<img src="/cgi-bin/mimetex.cgi?\tan{\alpha} = 2h_0/(a-b) = (2\cdot 60)/(100-40) = 2 \\ V(h)= l(h^2/\tan{\alpha} + bh) = l(h^2/2 + bh) \\ \Delta V(h) = l(h \Delta{h}+ b \Delta h) \\ \frac{\Delta V}{\Delta t} = l\left (h \frac{\Delta h}{\Delta t}+ b \frac{\Delta h}{\Delta t} \right ) \\ \frac{\Delta h}{\Delta t} = \frac{\Delta V}{\Delta t} \frac{1}{l(h+b)} = \frac{0.1}{5(0.2+0.4)} =0.0333 \ m/min" title="\tan{\alpha} = 2h_0/(a-b) = (2\cdot 60)/(100-40) = 2 \\ V(h)= l(h^2/\tan{\alpha} + bh) = l(h^2/2 + bh) \\ \Delta V(h) = l(h \Delta{h}+ b \Delta h) \\ \frac{\Delta V}{\Delta t} = l\left (h \frac{\Delta h}{\Delta t}+ b \frac{\Delta h}{\Delta t} \right ) \\ \frac{\Delta h}{\Delta t} = \frac{\Delta V}{\Delta t} \frac{1}{l(h+b)} = \frac{0.1}{5(0.2+0.4)} =0.0333 \ m/min" />

Denote the top side as a (a= 100 cm), the bottom side as b (b =40 cm),

the height of trapesoid as h0 (h0=60 cm), the height of the water as h (h = 20 cm) :

<img src="/cgi-bin/mimetex.cgi?\tan{\alpha} = 2h_0/(a-b) = (2\cdot 60)/(100-40) = 2 \\ V(h)= l(h^2/\tan{\alpha} + bh) = l(h^2/2 + bh) \\ \Delta V(h) = l(h \Delta{h}+ b \Delta h) \\ \frac{\Delta V}{\Delta t} = l\left (h \frac{\Delta h}{\Delta t}+ b \frac{\Delta h}{\Delta t} \right ) \\ \frac{\Delta h}{\Delta t} = \frac{\Delta V}{\Delta t} \frac{1}{l(h+b)} = \frac{0.1}{5(0.2+0.4)} =0.0333 \ m/min" title="\tan{\alpha} = 2h_0/(a-b) = (2\cdot 60)/(100-40) = 2 \\ V(h)= l(h^2/\tan{\alpha} + bh) = l(h^2/2 + bh) \\ \Delta V(h) = l(h \Delta{h}+ b \Delta h) \\ \frac{\Delta V}{\Delta t} = l\left (h \frac{\Delta h}{\Delta t}+ b \frac{\Delta h}{\Delta t} \right ) \\ \frac{\Delta h}{\Delta t} = \frac{\Delta V}{\Delta t} \frac{1}{l(h+b)} = \frac{0.1}{5(0.2+0.4)} =0.0333 \ m/min" />

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