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).

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).

Answer 1

`V_(B) = -2*\frac{(Xo + 2t)/(L) }{\sqrt{1-((Xo + 2t)/(L)) ^(2) } } \\ [m/s]`

Step-by-step explanation:

Since:

`Y_(B) = L*sin(\alpha )`

`cos(\alpha ) = (Xo + 2t)/(L) ---> \alpha = arcCos((Xo + 2t)/(L))` ---(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:

`d\alpha /dt = -\frac{1}{\sqrt{1-u^(2) } } *((2)/(L)) = -\frac{1}{\sqrt{1-cos\alpha ^(2) } }*((2)/(L))`

`d\alpha /dt = -(1)/(\sin\alpha ) *((2)/(L))`

Finally:

`V_(B) = L*cos\alpha *(-(1)/(sin\alpha )) *(2)/(L) = -2*Ctg\alpha [m/s]`

Or, in cartesians:

`V_(B) = -2*\frac{cos\alpha }{\sqrt{1-cos\alpha ^(2) } } [m/s]`

`V_(B) = -2*\frac{(Xo + 2t)/(L) }{\sqrt{1-((Xo + 2t)/(L)) ^(2) } } \\ [m/s]`

Answer 2

`(dy)/(dt)=-(2(7+2t))/(√(25^2-(7+2t)^2))\\`

for y (0,9)s

Step-by-step explanation:

the average velocity can be computed by using the derivative:

`(dy)/(dt)`

For this case we have

`y(t)=√(25^2-(7+2t)^2)=(25^2-(7+2t)^2)^{(1)/(2)}\\\\(dy)/(dt)=(1)/(2)(25^2-(7+2t)^2)^{-(1)/(2)}(-2(7+2t)(2))\\\\(dy)/(dt)=-(2(7+2t))/(√(25^2-(7+2t)^2))`

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

`7+2t\leq 25\\t\leq 9`

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

hope this helps!!