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\ Maria Vega Miguel Iglesias Edward Lara

Dario Yunes

Jose Baez

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What are related rates?Related rates are problems which involves at least two changing quantities and asks you to figure out the rate at which one is changing given sufficient information on all of the others. Usually the rate in common is time.>> 0 >> 1 >> 2 >> 3 >> 4 >>

Relative Rates Equationsdx ! Denotes the rate of change of x with respect to t. dt

Denotes the rate of change ofvolume with respect to t.

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Steps to solve related rates problemsStep 1. Read problem and make a sketch if possible. Step 2. Identify all variables, separate those that are given and those that have to be found. Step 3. Express all rates as derivatives. Step 4. Find an equation connecting variables. Step 5. Differentiate this equation using implicit differentiation. Step 6. Solve for the derivative that will give the unknown rate.>> 0 >> 1 >> 2 >> 3 >> 4 >>

Example 1

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Suppose that for a company making mini televisions , the cost and revenue equations given is C = 5,000 + 2x and R = 10x 0.001x2, Where the production output in 1 week is x mini televisions. If production is increasing at the rate of 500 mini televisions per week when production is 2,000 mini televisions, find the rate of increase in (a) Cost (b) Revenue

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Solution: These are really two related rates problems, one involving C, xand timet, and one involving R, x, and t. Differentiate the equations for C and R with respect to time.

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Cost is increasing at the rate of $1,000 per week.

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B)

R ! 10 x 0 . 001 x dR dx dx ! 10 0 . 002 x dt dt dt ! 10 500 0 . 002 2000 500 ! 3000

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Revenue is increasing at the rate of $3,000 per week.

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Exercise 2

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A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at 2 feet per second, how fast is the area changing when the radius is 10 feet? Use A = TR2

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Solution: Make a drawing. Given R= radius dR/dt= 2R= 10 Find dA/dtDifferentiate A = TR2. A= area.R

dA ! T 2 10 2 ! 40 T dt>> 0 >> 1 >> 2 >> 3 >> 4 >>

Exercise 3

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A balloon is being filled with air so that the radius is increasing at the rate 1in/2sec. How fast is the volume changing when the radius is 2.5 inches? V=4/3 r^3

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The Problem dv/dt = 4/3 * 3^2 dv/dt = 4/3 * 3^2 (simplify by eliminating the 3s) dv/dt = 4 r^2 dr/dt dv/dt=4 (2.5 inches)^2 * 1 inche/2 sec dv/dt= 39.25 in^3/sec

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A ladder 10 ft. long is leaning against a wall. If the bottom of the ladder slides away from the wall at 0.5 ft/sec. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft. from the wall.

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X2 + y2= 100 2xdx/dt + 2ydy/dt = 0 Ydy/dt = -xdx/dt Y=100-62 = 100-36 = 8 Substitute Y =8 8dy/dt = -6(0.5) 8dy/dt= -3 Dy/dt = -3/8 ft/sec

10 feet Y= dy/dt x= dx/dt

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