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Math Help - Calculus - rates of change

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    Senior Member topher0805's Avatar
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    Calculus - rates of change

    Two carts, A and B, are connected by a rope 43 ft long that passes over a pulley P (see the figure).



    The point Q is on the floor h = 14 ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 2 ft/s. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q?
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    Hello, topher0805!

    Two carts, A and B, are connected by a rope 43 ft long that passes over a pulley P.



    The point Q is on the floor; h = 14 ft directly beneath P and between the carts.
    Cart A is being pulled away from Q at a speed of 2 ft/s.
    How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q?
    Let x \,=\,AQ,\;y \,=\,QB

    In right triangle PQA\!:\;AP^2 \:=\:x^2 + 14^2\quad\Rightarrow\quad AP \:=\:\sqrt{x^2+196}

    In right triangle PQB\!:\;\;PB^2 \:=\:y^2+14^2\quad\Rightarrow\quad PB \:=\:\sqrt{y^2+196}


    The length of the rope is: . AP + PB \:=\:43

    . . So we have: . \left(x^2+196\right)^{\frac{1}{2}} + \left(y^2+196\right)^{\frac{1}{2}} \;=\;43\;\;{\color{blue}[1]}


    Differentiate with respect to time:

    . . \frac{1}{2}(x^2+196)^{\text{-}\frac{1}{2}}\!\cdot\!2x\!\cdot\!\frac{dx}{dt} \;+ \;\frac{1}{2}(y^2+196)^{\text{-}\frac{1}{2}}\!\cdot\!2y\!\cdot\!\frac{dy}{dt} \;\;=\;\;0

    Hence, we have: . \frac{dy}{dt} \;=\;-\frac{x}{y}\!\cdot\!\frac{\sqrt{y^2+196}}{\sqrt{x^  2+196}}\!\cdot\!\frac{dx}{dt}\;\;{\color{blue}[2]}


    We are given: . x = 5\text{ ft},\quad\frac{dx}{dt} = 2 ft/sec

    Substitute x = 5 into [1] and solve for y.

    Then substitute all that into [2].

    . . Bon voyage!

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