Math Problem Statement

A trough is 7 feet long and 11 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x4�=�4 from x=−1�=−1 to x=1�=1. The trough is full of water. Find the amount of work required to empty the trough by pumping the water over the top. 

Solution

This problem involves calculating the work required to pump water out of a trough. We can approach it by integrating the work done to lift infinitesimally small slices of water to the top of the trough.

Step 1: Define the situation

The trough has a length of 7 feet, a height of 11 feet, and the cross-sectional shape follows the curve y=x4y = x^4 from x=1x = -1 to x=1x = 1.

Let’s break down the steps:

  1. Volume of a slice of water: Consider a small horizontal slice of water at height yy, where the width of the slice is determined by the equation y=x4y = x^4. For a particular height yy, the horizontal distance between the two sides of the trough is the width, which is:

    Width=2(y1/4)\text{Width} = 2 \cdot (y^{1/4})

    The area of a cross-section at height yy is then:

    A(y)=Width×1=2(y1/4)A(y) = \text{Width} \times 1 = 2 \cdot (y^{1/4})

  2. Volume of the small slice: The volume of a small slice of water with thickness dydy is:

    dV=A(y)×7×dy=14(y1/4)dydV = A(y) \times 7 \times dy = 14 \cdot (y^{1/4}) \cdot dy

  3. Work done to lift the slice: The weight of a slice of water is ρgdV\rho \cdot g \cdot dV, where ρ\rho is the density of water (62.4 lb/ft³), and gg is the gravitational constant (in this case, the standard weight per volume is already given by ρg=62.4\rho g = 62.4).

    The height it must be lifted is 11y11 - y, since the trough is 11 feet tall. Therefore, the work to lift a slice is:

    dW=ρgdV(11y)=62.414(y1/4)(11y)dydW = \rho g \cdot dV \cdot (11 - y) = 62.4 \cdot 14 \cdot (y^{1/4}) \cdot (11 - y) \cdot dy

  4. Integrating the total work: To find the total work, we integrate from y=0y = 0 (the bottom of the trough) to y=1y = 1 (the top of the water):

    W=0162.414(y1/4)(11y)dyW = \int_0^1 62.4 \cdot 14 \cdot (y^{1/4}) \cdot (11 - y) \, dy

Let me compute this integral.import sympy as sp

Define the variable

y = sp.symbols('y')

Define the integrand: 62.4 * 14 * (y**(1/4)) * (11 - y)

integrand = 62.4 * 14 * (y**(1/4)) * (11 - y)

Compute the integral from y = 0 to y = 1

work = sp.integrate(integrand, (y, 0, 1)) workIt appears that the integral involves complex terms that SymPy cannot handle directly. Let me evaluate this integral numerically to get the result for the work done.import scipy.integrate as integrate import numpy as np

Define the function for the integrand

def integrand(y): return 62.4 * 14 * (y**(1/4)) * (11 - y)

Perform the numerical integration from y=0 to y=1

work, _ = integrate.quad(integrand, 0, 1) workIt seems like I can’t do more advanced data analysis right now. Please try again later. However, if you want, I can walk you through setting up the calculations manually or using a different method. Let me know how you'd like to proceed!

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Volume Calculations

Formulas

Integral of a product
Volume of a trough

Theorems

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Suitable Grade Level

Advanced High School