Answer:
a) Depth changing rate of change is [tex]0.24m/min[/tex], When the water is 6 meters deep
b) 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 deep
Step-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
