# Related Rate! P.GAO 高昊天 CHAD.BRINLEY 查德. Finding Related Rate The important Rule use of the CHAIN RULE. To find the rates of change of two or more related.

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Related Rate! P.GAO 高昊天 CHAD.BRINLEY 查德

Finding Related Rate The important Rule use of the CHAIN RULE. To find the rates of change of two or more related variables that are changing with respect to time!

HOW TO SOLVE THE PROBLEM! First, identify the rates of change we known and the rate of change will be found. Second, construct an equation relating the quantities which rates of change are known to the quantity that rate of change will be found. Third, differentiate both sides of the equation with respect to time.

Fourth, substitute the rates we known of change and the known quantities into the equation. Finally, solve for the wanted rate of change!

Example! A football is dropped from a height of 64 feet and at a horizontal distance of 16 feet from a light that is 64 feet above the ground at the top of a light pole. How fast is the shadow of the ball moving along the ground one second after the ball is dropped? Neglect air resistance so that the distance the football will have dropped as a function of time will be s = 16t 2 with the football dropped at t = 0. Can you draw in the right triangle with sides of lengths 16 and 16t 2 missing from the top of the picture below? You want to know the answer!?

ANSWER! Let x = the distance the football's shadow is from the base of the light pole. Using similar triangles we have x / 64 = 16 / 16t 2 x = 64t -2 dx/dt = -128t -3 dx/dt = -128 ft/sec

EXAMPLE A screen saver displays the outline of a 3 cm by 2 cm rectangle and then expands the rectangle in such a way that the 2 cm side is expanding at the rate of 4 cm/sec and the proportions of the rectangle never change. How fast is the area of the rectangle increasing when its dimensions are 12 cm by 8 cm?

Answer A = xy and 2 x = 3 y. dA/dt = [dx/dt] y + x [dy/dt] and 2 dx/dt = 3 dy/dt. Using these we see that dx/dt = (3/2)(4) = 6 cm/sec and that dA/dt = (6)(8) + (12)(4) = 96 cm 2 /sec.

EXAMPLE An 8 foot long ladder is leaning against a wall. The top of the ladder is sliding down the wall at the rate of 2 feet per second. How fast is the bottom of the ladder moving along the ground at the point in time when the bottom of the ladder is 4 feet from the wall.

ANSWER Y=the top of the ladder to the ground X=the bottom of the ladder to the wall dy/dt = -2 Find dx/dt when x = 4 x 2 + y 2 = 64 2x dx/dt + 2y dy/dt = 0 2x dx/dt - 4y = 0 dx/dt = 4y/(2x) = 2(3 1/2 ) ft/sec when x = 4 ft

That is it! And I’m going to move back china, thank you for giving me the wonderful time in America. I will come back to see you if I can! I hope all of us have a nice future! P.GAO

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