## anonymous one year ago I need help again! A 15-ft ladder is leaning against one wall of an alley 9 ft wide. the lade slips, its top sliding down the wall, its foot sliding across the alley and striking the opposite wall at a speed of 4ft/sec. how fast is the top of the ladder falling at that instance? calculus makes me want to cry!

1. Astrophysics

|dw:1438840509295:dw|

2. Astrophysics

Do you agree with the diagram

3. anonymous

i think so... i suck at calculus.

4. Astrophysics

|dw:1438841158267:dw| as it's sliding so we use the Pythagorean theorem $x^2+y^2=r^2$

5. anonymous

i got that part

6. Astrophysics

$x^2+y^2=15^2 \implies x^2+y^2=225$ now differentiate $2x \frac{ dx }{ dt }+2y \frac{ dy }{ dt } = 0$ Finish it off

7. Astrophysics

how fast is the top of the ladder falling at that instance this tells you what you're solving for, should be clear

8. Astrophysics

|dw:1438841472668:dw| just making sure you know it's a right triangle :P

9. anonymous

i just confused myself even further. I just can't figure this out.

10. Astrophysics

Ok, so given the equation what are we solving for?

11. anonymous

dt ??

12. Astrophysics

You are solving for dy/dt as we're looking for how fast the top of the ladder is falling

13. Astrophysics

So solve for dy/dt first, as you would for x, don't plug anything just isolate dy/dt.

14. anonymous

$\frac{ dy }{dt }=\frac{ -x \frac{ dx }{ dt } }{ y }$ ??

15. Astrophysics

Yeah :) good $\frac{ dy }{ dt }= -\frac{ x }{ y }\frac{ dx }{ dt }$

16. Astrophysics

So we know x = 9, and r = 15, what is y?

17. anonymous

12

18. Astrophysics

Good

19. Astrophysics

x = 9, y = 12, r = 15, and we know dx/dt = 4 ft/s $\frac{ dy }{ dt } = - \frac{ 9 }{ 12 }(4) ft/s$

20. anonymous

so 3ft/s ?

21. Astrophysics

- 3ft/s the negative just means the distance from the top of the ladder is decreasing

22. anonymous

oh okay i get it now!

23. anonymous

thank you so much

24. Astrophysics

No problem