# A 18 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 2ft/sec.

Find the velocity of the top of the ladder at time t=2

## Dude.... You're so smart.... Wow.

## To find the velocity of the top of the ladder at time t=2, we need to use the properties of similar triangles and apply the chain rule of differentiation.

Let's denote the distance of the top of the ladder from the wall as y(t), where t represents time.

According to the given information, the bottom of the ladder is 3 ft from the wall at time t=0, and it slides away from the wall at a rate of 2ft/sec. This means that the distance of the bottom of the ladder from the wall can be represented as x(t) = 3 + 2t.

Now, we can form a right triangle involving the ladder, the distance from the wall, and the height on the wall:

/|

y(t)/ |

/ |

/ |

/ |

θ / |

/ |

/____|

Using the properties of similar triangles, we can relate the height on the wall to the distance of the bottom of the ladder from the wall:

y(t) / x(t) = 18 / 20

Simplifying the equation, we have:

y(t) = (18/20) * x(t)

y(t) = (9/10) * x(t)

Now, we need to differentiate both sides of the equation with respect to time t using the chain rule.

(dy(t) / dt) = (9/10) * (dx(t) / dt)

The velocity of the top of the ladder is given by (dy(t) / dt), and we need to find its value at time t=2.

So, let's calculate the derivative of the distance of the bottom of the ladder from the wall.

dx(t) / dt = 2 ft/sec

Substituting this value into our equation, we get:

(dy(t) / dt) = (9/10) * 2 ft/sec

(dy(t) / dt) = (9/5) ft/sec

Therefore, the velocity of the top of the ladder at time t=2 is (9/5) ft/sec.

## To find the velocity of the top of the ladder at time t=2, we can use the Pythagorean theorem to relate the distance the bottom of the ladder moves away from the wall, the height of the ladder, and the distance the top of the ladder slides down the wall.

Let's assume that the distance the top of the ladder has slid down the wall at time t=0 is x.

Using the Pythagorean theorem, we have:

(3 ft + 2t)^2 = x^2 + (18 ft)^2

Simplifying the equation:

9 + 12t + 4t^2 = x^2 + 324

Taking the derivative of both sides with respect to t:

24t + 12 = 2x(dx/dt)

We want to find the velocity of the top of the ladder at t=2, which means we need to find dx/dt when t=2.

Substituting t=2 into the equation:

24(2) + 12 = 2x(dx/dt) at t=2

48 + 12 = 2x(dx/dt) at t=2

Combining like terms:

60 = 2x(dx/dt) at t=2

Dividing both sides by 2x:

(dx/dt) = 60 / (2x) at t=2

Now we need to find the value of x when t=2. Plugging t=2 into the equation (3 ft + 2t), we get:

x = 3 + 2(2)

= 3 + 4

= 7 ft

Substituting x=7 into the equation:

(dx/dt) = 60 / (2(7)) at t=2

Simplifying:

(dx/dt) = 60 / 14

≈ 4.29 ft/sec

Therefore, the velocity of the top of the ladder at time t=2 is approximately 4.29 ft/sec.

## This is the standard question in just about every textbook when introducing rates of change.

Did you make your diagram?

I see a right angled triangle.

At a time of t seconds, let the foot of the ladder be x ft from the wall, and let the top of the ladder be y ft above ground.

Clearly,

x^2 + y^2 = 18^2

differentiate with respect to t

2x dx/dt + 2y dy/dt = 0

at t = 2, x = 7,

49 + y^2 = 18^2

y = √275 or 5√11

sub in all we know,

2(7)(2) + 2(5√11) (dy/dt) = 0

dy/dt = -28/(10√11) ft/s

= appr .844 ft/sec

check my arithmetic