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A water trough is 10ft long and its ends are isosceles triangles that are 3ft across the top and have height 1ft. If [tex]\frac{dV}{dt}=12\frac{\mbox{ft}^{3}}{\mbox{min}}[/tex], how fast is the water level rising when the water is 6 inches deep?

[tex]V=\frac{1}{2}(b_1 + b_2)h \rightarrow \frac{1}{2}[b_1 + (b_1 + 2r)]h[/tex]

... By similar triangles [tex]\frac{3}{1}=\frac{r}{h} \rightarrow r = 3h[/tex]

[tex]V=\frac{1}{2}(20+6h)h \rightarrow (10+3h)h[/tex]

[tex]\frac{dV}{dt}=(10+6h)\frac{dh}{dt}[/tex]

Is this correct so far?

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# Homework Help: Related Rates - Set up

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