Answer to Question #243706 in Calculus for CELL

Question

TOPIC: General Application of Derivatives. Draw the necessary figure and indicate the dimension given.

• A 5m ladder leans against a vertical wall. If the top starts sliding downward at the rate of 5.0 ft/sec, find how fast in m/sec the lower end moves when it is 4 m from the wall.

Answer 1

By the Pythagorean Theorem

$x^2+y^2=L^2$

Differentiate both sides with respect to $t$ and use the Chain Rule

$\dfrac{d}{dt}(x^2+y^2)=\dfrac{d}{dt}(L^2)$

$2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$

$\dfrac{dx}{dt}=-\dfrac{y}{x}\cdot\dfrac{dy}{dt}$

Given $\dfrac{dy}{dt}=-5.0\ ft/sec.$

When $x=4\ m$

$y=\sqrt{L^2-x^2}$

$y=\sqrt{(5\ m)^2-(4\ m)^2}=3\ m$

$\dfrac{y}{x}=\dfrac{3\ m}{4\ m}=0.75$

Substitute

$\dfrac{dx}{dt}=-0.75\cdot(-5.0\ ft/sec)=3.75\ ft/sec$

$3.75\ ft/sec=3.75(0.3048)\ m/sec=1.143 \ m/sec$

The lower end moves right at the rate of $1.143 \ m/sec.$