3. a water trough is 8 m long and has a cross - section of an isosceles triangle and is 80 cm across and 50…

3. a water trough is 8 m long and has a cross - section of an isosceles triangle and is 80 cm across and 50 cm high. the trough is being filled with water at a rate of 0.3 m³/min. how fast is the water level rising when the water is 30 cm deep? ($v=\frac{1}{2}whl$)

3. a water trough is 8 m long and has a cross - section of an isosceles triangle and is 80 cm across and 50 cm high. the trough is being filled with water at a rate of 0.3 m³/min. how fast is the water level rising when the water is 30 cm deep? ($v=\frac{1}{2}whl$)

Answer

Explanation:

Step1: Relate variables

Let the water - level be $h$ and the width of the water - surface be $w$. Since the cross - section is an isosceles triangle, by similar triangles, $\frac{w}{h}=\frac{80}{50}=\frac{8}{5}$, so $w=\frac{8}{5}h$. The length of the trough $L = 8m=800cm$. The volume of water $V$ in the trough is $V=\frac{1}{2}whL$. Substituting $w = \frac{8}{5}h$ and $L = 800$ into the volume formula, we get $V=\frac{1}{2}\times\frac{8}{5}h\times h\times800=640h^{2}$.

Step2: Differentiate with respect to time

Differentiate both sides of the equation $V = 640h^{2}$ with respect to time $t$ using the chain - rule. $\frac{dV}{dt}=1280h\frac{dh}{dt}$.

Step3: Solve for $\frac{dh}{dt}$

We know that $\frac{dV}{dt}=0.3m^{3}/min = 300000cm^{3}/min$ and $h = 30cm$. Substitute these values into the equation $\frac{dV}{dt}=1280h\frac{dh}{dt}$. Then $300000=1280\times30\times\frac{dh}{dt}$. First, simplify the right - hand side: $1280\times30 = 38400$. So, $\frac{dh}{dt}=\frac{300000}{38400}=\frac{125}{16}=7.8125cm/min$.

Answer:

$7.8125cm/min$