A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and height is 10 meters. Assume that the two equal length sides of the triangle are the sides of the water tank and the other side of the triangle is the top of the tank and is parallel to the ground. Water is being pumped into the tank at a rate of 2 m3/min. When the water is 6 meters deep at what rate is (a) depth changing and (b) the width of the top of the water changing? Note the sketch below is not to scale…

Answer:a) Depth changing rate of change is [tex]0.24m/min[/tex], When the water is 6 meters deepb) The width of the top of the water is changing at a rate of [tex]0.17m/min[/tex], When the water is 6 meters deepStep-by-step explanation:As we can see in the attachment part II, there are similar triangles, so we have the following relation between them [tex]\frac{3.5}{10} =\frac{a}{h}[/tex], then [tex]a=0.35h[/tex].a) As we have that volume is [tex]V=\frac{1}{2} 2ahL=ahL[/tex], then [tex]V=(0.35h^{2})L[/tex], so we can derivate it [tex]\frac{dV}{dt}=2(0.35h)L\frac{dh}{dt}[/tex] due to the chain rule, then we clean this expression for [tex]\frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt}[/tex] and compute with the knowns [tex]\frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min[/tex], is the depth changing rate of change when the water is 6 meters deep.b) As the width of the top is [tex]2a=0.7h[/tex], we can derivate it and obtain [tex]\frac{da}{dt}=0.7\frac{dh}{dt}  =0.7*0.24m/min=0.17m/min[/tex] The width of the top of the water is changing, When the water is 6 meters deep at this rate