Questions: A trough has ends shaped like isosceles triangles, and has a length of 10 m, width 3 m, and height 4 m. Water is being pumped into the trough at a rate of 1 m^3 / min. At what rate does the height of the water change when the water is 3 m deep? dh/dt m / min

A trough has ends shaped like isosceles triangles, and has a length of 10 m, width 3 m, and height 4 m. Water is being pumped into the trough at a rate of 1 m^3 / min. At what rate does the height of the water change when the water is 3 m deep?

dh/dt m / min

Transcript text: A trough has ends shaped like isosceles triangles, and has a length of 10 m , width 3 m , and height 4 m. Water is being pumped into the trough at a rate of $1 \mathrm{~m}^{3} / \mathrm{min}$. At what rate does the height of the water change when the water is 3 m deep? $\frac{d h}{d t}$ $\square$ $\mathrm{m} / \mathrm{min}$

Solution

Solution Steps

Step 1: Understand the Problem

We are given a trough with isosceles triangle ends, a length of 10 ft, width of 3 ft, and height of 4 ft. Water is being pumped into the trough at a rate of 12 ft³/min. We need to find the rate at which the height of the water is changing when the water is 3 ft deep.

Step 2: Set Up the Volume Formula

The volume $ V $ of the trough can be expressed as: \[ V = \text{Area of cross-section} \times \text{Length} \] The cross-section is an isosceles triangle with base $ b $ and height $ h $. The area $ A $ of the triangle is: \[ A = \frac{1}{2} b h \] Since the base and height of the triangle change with the water level, we need to express $ b $ in terms of $ h $.

Step 3: Relate Base and Height

Given the dimensions of the trough, when the height $ h $ is 4 ft, the base $ b $ is 3 ft. The relationship between $ b $ and $ h $ is linear: \[ \frac{b}{h} = \frac{3}{4} \] So, for any height $ h $: \[ b = \frac{3}{4} h \]

Step 4: Express Volume in Terms of Height

Substitute $ b = \frac{3}{4} h $ into the area formula: \[ A = \frac{1}{2} \left( \frac{3}{4} h \right) h = \frac{3}{8} h^2 \] Then, the volume $ V $ is: \[ V = \left( \frac{3}{8} h^2 \right) \times 10 = \frac{30}{8} h^2 = \frac{15}{4} h^2 \]

Step 5: Differentiate with Respect to Time

Differentiate both sides of the volume equation with respect to time $ t $: \[ \frac{dV}{dt} = \frac{15}{4} \cdot 2h \frac{dh}{dt} = \frac{15}{2} h \frac{dh}{dt} \]

Step 6: Substitute Known Values

We know $ \frac{dV}{dt} = 12 $ ft³/min and $ h = 3 $ ft. Substitute these values into the differentiated equation: \[ 12 = \frac{15}{2} \cdot 3 \cdot \frac{dh}{dt} \] \[ 12 = \frac{45}{2} \cdot \frac{dh}{dt} \] \[ 12 = 22.5 \cdot \frac{dh}{dt} \]

Step 7: Solve for $ \frac{dh}{dt} $

\[ \frac{dh}{dt} = \frac{12}{22.5} = \frac{4}{7.5} = \frac{8}{15} \]