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).

I found the parametric equation for y(t) =√ 25^2 -(7+2t)^2

Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places).

for y (0,9)s

Explanation:

the average velocity can be computed by using the derivative:

For this case we have

However, for a real result the values of t must be

that is, an allowed interval for t is (0,9)

hope this helps!!

Explanation:

Since:

---(1)

By compositive derivate:

d(YB)/dt =VB = L*(d[sin(α)]/dα)*(dα/dt) = L*cos(α)*dα/dt --- (2)

From the expresion (1):

if we call; u = (Xo + 2t)/L

Then: dα/dt = (dα/du)*(du/dt) = (d[arcCos(u)]/du)*(d[(Xo + 2t)/L]/dt)

Thus:

Finally:

Or, in cartesians:

object 2 because all the other ones are wrong

seeing the graph would