A 13-ft ladder is leaning against a house when its base starts to slide away. By

Mahagnazk

Mahagnazk

Answered question

2021-11-21

A 13-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate of 5 ft/sec. 
a. How fast is the top of the ladder sliding down the wall then? 
b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? 
c. At what rate is the angle u between the ladder and the ground changing then?

Answer & Explanation

Volosoyx

Volosoyx

Beginner2021-11-22Added 10 answers

We are aware that the ladder is a certain length l=13 ft 
We also know that for x=12, the rate of transformation is  dx  dt =5 
From the given information, we can infer that for a fixed t,l=13=x2+y2 
a. We want to find the speed of the ladder sliding the wall when x=12, which is represented by  dy  dt 
From the given equation 132=x2+y2, we can find the speed: 
x2+y2=132 
2x dx  dt +2y dy  dt =0 
Since for x=12, y=132122=5, we have: 
2x dx  dt +2y dy  dt =0 
2125+25 dy  dt =0 
 dy  dt =12 ftsec  
b. We want to find the rate of change dA dt  of the area A of th triangle formed by the ladder, wall and the ground. Equation of the area is: 
A=xy2=x132x22 
We already know that  dx  dt =5 
 

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