Word problem ( calculus 1)

Question

A 13 foot ladder is leaning against a vertical wall. If the top of the ladder slips down the wall at a rate of 2ft/s.

 

how fast will the base of the ladder be moving away from the wall when the top of the ladder is 5 feet above the ground ?

 

 

how fast the angle between the ladder and the ground decreasing ?

i got 5/5ft/sec for the first part and the second part I got 1/6 radians /sec. please help me check my work! 

Answer 1

This is a related rate problem. A sketch will be helpful. The ladder, vertical wall, and the ground will form a right triangle. If we call the distance from the top of the ladder to the ground v, then dv/dt = -2 ft/s. Now, let h = the distance from the base of the wall to the bottom of the ladder. Our task is to find dh/dt when v = 5 ft.

 

Using the Pythagorean Theorem, we know that v² + h² = 13². The leaning ladder forms the hypotenuse of the right triangle. Now, let's differentiate both sides of this equation with respect to t.

 

2v (dv/dt) + 2h (dh/dt) = 0

 

2h (dh/dt) = -2v (dv/dt)

 

dh/dt = -v/h (dv/dt)

 

We know the value of dv/dt is -2 ft/s, and the value of v = 5 ft. Since the ladder has a fixed length of 13 ft and we know that at the moment in question v = 5, we can use the Pythagorean Theorem to find h. 

 

5² + h² = 13²

 

25 + h² = 169

 

h² = 144

 

h = 12

 

Now, we can compute dh/dt.

 

dh/dt = (-5/12)(-2) = 5/6 ft/s

 

You can use one of the trig ratios to solve the second portion of the problem.