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Sagot :
Answer:
Approximately [tex]0.033\; {\rm m\cdot min^{-1}}[/tex] (meters per minute.)
Explanation:
Let [tex]A[/tex] and [tex]B[/tex] denote the bottom and top width of the trough, respectively. It is given that [tex]A = 30\; {\rm cm} = 0.30\; {\rm m}[/tex] and [tex]B = 80\; {\rm cm} = 0.80\; {\rm m}[/tex]. Let [tex]H[/tex] denote the height of this trough; [tex]H = 50\; {\rm cm} = 0.50\; {\rm m}[/tex].
Let [tex]h[/tex] denote the current depth of the water in this trough.
Let [tex]b[/tex] denote the current width of the surface of the water. As water fills the trough, this width increases from [tex]A = 0.30\; {\rm m}[/tex] (width of bottom of trough) to [tex]B = 0.80\; {\rm m}[/tex] (width of top of trough.)
The relationship between [tex]b[/tex] and [tex]h[/tex] is linear:
[tex]\displaystyle b = \frac{h}{H}\, (B - A) + A[/tex].
Cross-section area of water in this trough:
[tex]\begin{aligned}(\text{area}) &= \frac{1}{2}\, (A + b) \, h \\ &= \frac{1}{2}\, \left(A + \frac{h}{H}\, (B - A) + A\right)\, h \\ &= A\, h + \frac{1}{2\, H} (B - A)\, h^{2}\end{aligned}[/tex].
Let [tex]L[/tex] denote the length of this trough; [tex]L = 10\; {\rm m}[/tex].
Let [tex]v[/tex] denote the volume of water in this trough:
[tex]\begin{aligned}v &= (\text{area})\, L \\ &= \frac{1}{2}\, (A + b) \, h\, L \\ &= \frac{1}{2}\, \left(A + \frac{h}{H}\, (B - A) + A\right)\, h\, L \\ &= A\, L\, h + \frac{L}{2\, H} (B - A)\, h^{2}\end{aligned}[/tex].
Differentiate both sides implicitly with respect to [tex]v[/tex]:
[tex]\displaystyle \frac{d}{dv}[v] = \frac{d}{dv}\left[A\, L\, h + \frac{L}{2\, H}\, (B - A)\, h^{2}\right][/tex].
[tex]\displaystyle \frac{d}{dv}[v] = \frac{d}{dv}[A\, L\, h] + \frac{d}{dv}\left[\frac{L}{2\, H}\, (B - A)\, h^{2}\right][/tex].
[tex]\displaystyle 1 = A\, L\, \frac{dh}{dv} + \frac{L}{2\, H}\, (B - A)\, 2\, h\, \frac{dh}{dv}[/tex].
(Note that [tex]A[/tex], [tex]B[/tex], [tex]L[/tex], and [tex]H[/tex] are constants.)
Rearrange this equation to obtain:
[tex]\displaystyle 1 = A\, L\, \frac{dh}{dv} + \frac{L}{H}\, (B - A)\, h\, \frac{dh}{dv}[/tex].
[tex]\displaystyle 1 = \left(A\, L + \frac{L}{H}\, (B - A)\, h \right)\, \frac{dh}{dv}[/tex].
[tex]\displaystyle \frac{dh}{dv} = \frac{1}{\displaystyle A\, L + \frac{L}{H}\, (B - A)\, h}[/tex].
Let [tex]t[/tex] denote time. It is given that the trough is being filled at a rate of [tex]0.2\; {\rm m^{3}\cdot min^{-1}}[/tex]. In other words:
[tex]\displaystyle \frac{dv}{dt} = 0.2\; {\rm m^{3}\cdot min^{-1}}[/tex].
Apply the chain rule to find the rate at which [tex]h[/tex] is changing with respect to time [tex]t[/tex]:
[tex]\begin{aligned} \frac{dh}{dt} &= \frac{dh}{dv}\cdot \frac{dv}{dt} \\ &= \frac{1}{\displaystyle A\, L + \frac{L}{H}\, (B - A)\, h}\cdot \frac{dv}{dt} \end{aligned}[/tex].
Substitute in [tex]A = 0.30\; {\rm m}[/tex], [tex]L = 10\; {\rm m}[/tex], [tex]H = 0.50\; {\rm m}[/tex], [tex]B = 0.80\; {\rm m}[/tex], [tex]h = 0.30\; {\rm m}[/tex] (converted from [tex]30\; {\rm cm}[/tex]), and that the rate of change in [tex]v[/tex] is [tex]0.2\; {\rm m^{3}\cdot min^{-1}}[/tex]:
[tex]\begin{aligned} \frac{dh}{dt} &= \frac{dh}{dv}\cdot \frac{dv}{dt} \\ &= \frac{1}{\displaystyle A\, L + \frac{L}{H}\, (B - A)\, h}\cdot \frac{dv}{dt} \\ &= \frac{0.2\; {\rm m^{3}\cdot min^{-1}}}{\displaystyle 0.30\; {\rm m} \times 10\; {\rm m} + \frac{10\; {\rm m}}{0.60\; {\rm m}}\, (0.80\; {\rm m} - 0.30\; {\rm m}) \, 0.30\; {\rm m}} \\ &=0.033\; {\rm m\cdot min^{-1}} \end{aligned}[/tex].
In other words, the depth of the water in this trough increases at a rate of approximately [tex]0.033\; {\rm m \cdot min^{-1}}[/tex] (meters per minute.)
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