Respuesta :
Answer:
[tex]\dfrac{dh}{dt}=21 \text{m/min}[/tex]
the rate of change of height when the water is 1 meter deep is 21 m/min
Step-by-step explanation:
First we need to find the volume of the trough given its dimensions and shape: (it has a prism shape so we can directly use that formula OR we can multiply the area of its triangular face with the length of the trough)
[tex]V = \dfrac{1}{2}(bh)\times L[/tex]
here L is a constant since that won't change as the water is being filled in the trough, however 'b' and 'h' will be changing. The equation has two independent variables and we need to convert this equation so it is only dependent on 'h' (the height of the water).
As its an isosceles triangle we can find a relationship between b and h. the ratio between the b and h will be always be the same:
[tex]\dfrac{b}{h} = \dfrac{5}{7}[/tex]
[tex]b=\dfrac{5}{7}h[/tex] this can be substituted back in the volume equation
[tex]V = \dfrac{5}{14}h^2L[/tex]
the rate of the water flowing in is:
[tex]\dfrac{dV}{dt} = 6[/tex]
The question is asking for the rate of change of height (m/min) hence that can be denoted as: [tex]\frac{dh}{dt}[/tex]
Using the chainrule:
[tex]\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}[/tex]
the only thing missing in this equation is dh/dV which can be easily obtained by differentiating the volume equation with respect to h
[tex]V = \dfrac{5}{14}h^2L[/tex]
[tex]\dfrac{dV}{dh} = \dfrac{5}{7}hL[/tex]
reciprocating
[tex]\dfrac{dh}{dV} = \dfrac{7}{5hL}[/tex]
plugging everything in the chain rule equation:
[tex]\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}[/tex]
[tex]\dfrac{dh}{dt}=\dfrac{7}{5hL}\times 6[/tex]
[tex]\dfrac{dh}{dt}=\dfrac{42}{5hL}[/tex]
L = 12, and h = 1 (when the water is 1m deep)
[tex]\dfrac{dh}{dt}=\dfrac{42}{5(1)(12)}[/tex]
[tex]\dfrac{dh}{dt}=21 \text{m/min}[/tex]
the rate of change of height when the water is 1 meter deep is 21 m/min
The rate at which the height of the water changes when the water is 1 m deep is; dh/dt = 8.4 m/min
We are given;
Width; w = 5 m
Height; h = 7 m
Length of trough; L = 12 m
This means that; w = ⁵/₇h
The formula for Volume of the trough is:
V = Area of triangle × length
V = ¹/₂whL
V = ¹/₂ × ⁵/₇h × h × 12
V = ³⁰/₇ h²
Differentiating using chain rule for both sides with respect to t gives;
dV/dt = (⁶⁰/₇)h(dh/dt) .
Now, we are given dV/dt = 6 m³/min and we want to find the rate at which the height of the water changes when the water is 1 m deep.
From similar triangles, we can say that h = ¹/₁₂ m. Thus;
6 = (⁶⁰/₇)(¹/₁₂)(dh/dt)
⁵/₇(dh/dt) = 6
dh/dt = 42/5
dh/dt = 8.4 m/min
Read more at; https://brainly.com/question/15690642
