Math Problem Statement

A winch at the top of a 12-meter building pulls a pipe of the same length to a vertical position, as shown in the figure. The winch pulls in rope at a rate of −0.4 meter per second (ds/dt = −0.4 m/sec) . Find the rate of vertical change and the rate of horizontal change at the end of the pipe when y = 4. (Round your answers to three decimal places.)

Solution

Let's break down the problem step by step.

Step 1: Define the scenario and variables.

We have a winch at the top of a 12-meter building pulling a 12-meter pipe. The pipe is being pulled in by the winch, so the winch reduces the length of the rope at a rate of dsdt=0.4m/sec\frac{ds}{dt} = -0.4 \, \text{m/sec}.

  • ss is the length of the rope being pulled in.
  • xx is the horizontal distance from the bottom of the pipe to the base of the building.
  • yy is the vertical height from the bottom of the pipe to the ground.

Given:

  • The total length of the pipe is 12 meters, so by the Pythagorean theorem: s2=x2+y2s^2 = x^2 + y^2 Since the pipe remains 12 meters in length, the equation becomes: 122=x2+y212^2 = x^2 + y^2 Simplifying: 144=x2+y2144 = x^2 + y^2

  • The rate at which the rope is being pulled is dsdt=0.4m/sec\frac{ds}{dt} = -0.4 \, \text{m/sec}.

  • We need to find the rate of vertical change (dydt\frac{dy}{dt}) and horizontal change (dxdt\frac{dx}{dt}) when y=4y = 4.

Step 2: Solve for xx when y=4y = 4.

From the equation 144=x2+y2144 = x^2 + y^2, when y=4y = 4, we get: x2+42=144x^2 + 4^2 = 144 x2+16=144x^2 + 16 = 144 x2=128x^2 = 128 x=128=8211.314x = \sqrt{128} = 8\sqrt{2} \approx 11.314

Step 3: Differentiate the equation x2+y2=144x^2 + y^2 = 144.

To find the rates of change, differentiate both sides of the equation with respect to time tt: 2xdxdt+2ydydt=02x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 This simplifies to: x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \tag{1}

Step 4: Relating ds/dtds/dt to dx/dtdx/dt and dy/dtdy/dt.

Since s2=x2+y2s^2 = x^2 + y^2 and s=12s = 12, we can differentiate s=x2+y2s = \sqrt{x^2 + y^2} with respect to tt: dsdt=12s(2xdxdt+2ydydt)\frac{ds}{dt} = \frac{1}{2s}(2x \frac{dx}{dt} + 2y \frac{dy}{dt}) Simplifying: dsdt=xdxdt+ydydts\frac{ds}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{s} Given that s=12s = 12 and dsdt=0.4\frac{ds}{dt} = -0.4, we have: -0.4 = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{12} \tag{2}

Step 5: Solve for dydt\frac{dy}{dt} and dxdt\frac{dx}{dt}.

Now, we have two equations:

  • Equation (1): xdxdt+ydydt=0x \frac{dx}{dt} + y \frac{dy}{dt} = 0
  • Equation (2): 0.4=xdxdt+ydydt12-0.4 = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{12}

Substitute xdxdt+ydydt=12×(0.4)x \frac{dx}{dt} + y \frac{dy}{dt} = 12 \times (-0.4): xdxdt+ydydt=4.8x \frac{dx}{dt} + y \frac{dy}{dt} = -4.8

From Equation (1), express dxdt\frac{dx}{dt} in terms of dydt\frac{dy}{dt}: dxdt=yxdydt\frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt} Substitute this into Equation (2): x(yxdydt)+ydydt=4.8x \left(-\frac{y}{x} \frac{dy}{dt}\right) + y \frac{dy}{dt} = -4.8 Simplify: ydydt+ydydt=4.8-y \frac{dy}{dt} + y \frac{dy}{dt} = -4.8 Since the terms involving dydt\frac{dy}{dt} cancel out, we are left with:

Thus, the rate of

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Math Problem Analysis

Mathematical Concepts

Related Rates
Differentiation
Pythagorean Theorem

Formulas

s^2 = x^2 + y^2
2x(dx/dt) + 2y(dy/dt) = 0
ds/dt = (x(dx/dt) + y(dy/dt)) / s

Theorems

Pythagorean Theorem
Chain Rule

Suitable Grade Level

Grades 11-12 (High School Calculus)