Answer to Question #183862 in Calculus for Chayzel Cayanga

Question

1. A baseball diamond has the shape of a square with sides 90 ft long. A player 60 ft from second base is running towards third base at a speed of 28 ft/min. At what rate is the player’s distance from the home plate changing?
2. A ladder inclined at 60 ° with the horizontal is leaning against a vertical wall. The foot of the ladder is 3 meters away from the foot of the wall. A boy climbs the ladder such that his distance z meters with respect to the foot of the ladder is given by z=6t, where t is the time in seconds. Find the rate at which his vertical distance from the ground changes with respect to t. Find the rate at which his distance from the foot of the wall is changing with respect to t when he is 3 meter away from the foot of the ladder.

Answer 1

From the figure above,

s² = x² + 90² -----(i)

To get ds/dt, we will find the relation between ds/dt and dx/dt by differentiating equation (1) with respect to t as follows.

$\dfrac{d}{dt}s² =\dfrac{d}{dt}(x²+9²)$ (differentiating both sides w.r.t. t)

$\dfrac{d}{dt}s² =\dfrac{d}{dt}(x²+0)$ (sum and constant rules)

$2s\dfrac{ds}{dt} = 2x\dfrac{dx}{dt}$ (general power rule)

$\dfrac{ds}{dt} = \dfrac{2x}{2s}\dfrac{dx}{dt}$ (dividing both sides by 2s)

$\dfrac{ds}{dt} =\dfrac{x}{\sqrt{x²+90²}} \dfrac{dx}{dt}$ (from equation (1))

$\dfrac{ds}{dt} = \dfrac{20}{\sqrt{x²+90²}}.25 = \dfrac{-50}{\sqrt{85}} \ ft/sec$ (getting the needed rate)

Note that the rate of the change is negative because it decreases.

$\dfrac zx= \tan60°$

$z = x\tan60° \\ z=x√3$

$\dfrac{dz}{dt}= \dfrac{dx}{dt}\sqrt3$

$6\ m/s= \dfrac{dx}{dt}\sqrt3$

$\dfrac{dx}{dt}= \dfrac{6}{\sqrt3} = 3.46\ m/s$