the top of a 17 foot ladder

Question

The top of a 17 foot ladder, leaning against a vertical wall, is slipping down the wall at the rate of 4 feet per second. How fast is the bottom of the ladder sliding along the ground when the bottom of the ladder is 11 feet away from the base of the wall?

Answer 1

 

X represents the measure from the bottom of the ladder to the base of the wall:  horizontal distance

 

Y represents the measure from the top of the ladder to the the base of the wall : vertical distance.

 

X and Y BOTH functions of time.  

 

Hence a right triangle forms by the ladder, wall, and ground with the ladder as the hypotenuse.

 

By pythagorean theorem, when the base of the ladder is 11 feet away from the base of the wall,

     11^2 + y^2 = 17^2

 

    y^2 = 289 - 121 = 168. So  the top of the ladder is square-root(168) feet from the base of the ground

when the bottom of the ladder is 11 feet away from the base of the ground. 

 

When you prime factor 168, Square-root(168) = 2*square-root(42), which I will abbreviate as 2*rad42 henceforth.

 

So x(11) implies that y(11) = 2*rad42

 

In general, pythagoream theorem guarantees 289 = x^2 + y^2

 

The derivative is :  0 = 2*X*dX/dt  + 2*Y*dY/dt

 

                         

                           0 = X*dX/dt +  Y*dY/dt   <--- 2 factors out and cancels

 

   Now plugging in the given and calculated values, as we are trying to find dX/dT

 

                          0 = 11* dX/dt + 2*rad42*4     ( dY/dt given as 4, Y==2*rad42 when x=11)

 

                      dX/dt = -8*rad42/11 = -4.713266

 

It is negative because it is moving away from the wall, leftwards towards negative infinity