Question

A ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall. The foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The location of the foot of the ladder at time t seconds is given by the parametric equations (7+2t,0).

Answer 1

The location of the foot of the ladder at time t is $y=\sqrt{625-(7+2 t)^{2}}$

Step-by-step explanation:

The wall, ladder and the ground makes a right angled triangle. Let the ground be x and the wall be y.

Then, by Pythagorean theorem, we have,

$x^{2} +y^{2} =25^{2} \\x^{2} +y^{2} =625$

Also, given that the ladder is sliding at a rate of 2ft/sec.

The location of the foot of the ladder at time t is given by the parametric equations $(7+2t,0)$ and thus, $x=7+2t$. Substituting these values in the equation $x^{2} +y^{2} =625$, we get,

$x^{2} +y^{2} =625\\(7+2t)+y^{2} =625$

Subtracting $7+2t$ on both sides,

$y^{2} =625-(7+2t)$

Taking square root on both sides, we get,

$y=\sqrt{625-(7+2 t)^{2}}$

Thus, the location of the foot of the ladder at time t is $y=\sqrt{625-(7+2 t)^{2}}$