# Thread: Calculus - rates of change

1. ## 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?

2. 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 $\displaystyle x \,=\,AQ,\;y \,=\,QB$

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

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

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

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

Differentiate with respect to time:

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

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

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

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

Then substitute all that into [2].

. . Bon voyage!