Help!!! A 25 ft ladder is leaning against a vertical wall. The bottom of the ladder is pulled horizontally away from the wall at 3 ft/sec. Determine how fast the top of the ladder is sliding when the bottom of the ladder is 15 ft from the wall.

A) –4 ft/sec
B) –2.25 ft/sec
C) –13.375 ft/sec
D)–12.25 ft/sec
E) –0.75 ft/sec

Question

Help!!! A  25 ft ladder is leaning against a vertical wall.  The bottom of the ladder is pulled horizontally away from the wall at 3 ft/sec. Determine how fast the top of the ladder is sliding when the bottom of the ladder is 15 ft from the wall.

A)  –4 ft/sec
B)  –2.25 ft/sec	
C) –13.375  ft/sec
D)–12.25 ft/sec
E)  –0.75 ft/sec

anonymous · Answer

Let x be the horizontal distance from the wall and y be the point where the ladder touches the wall. Then $$x^2+y^2=25^2.$$

Then the rate at which the top of the ladder is sliding is $$dy/dt.$$

To find $$dy/dt$$, differentiate the relation $$x^2+y^2=25^2.$$ implicitly with respect to $$t.$$

We are given that $$dx/dt=3$$ and $$x=15$$. From the relation, we can find $$y$$ when $$x=15.$$

I don't want to give away the answer, but just differentiate implicitly w.r.t. t and solve for dy/dt. Good luck!