A 26 foot ladder is placed againts a wall.If the top of the ladder is sliding down at a rate of 2 feet per second,then estimate the rate at which the bottom of the ladder is moving away from the wall when the bottom of the ladder is 10feet away from the wall.

Let height and base be h and b respectively.All I could understand was that we have to find db/dt,we are given dh/dt ..can somebody help me with this question?

2. Originally Posted by mathcalculushelp
A 26 foot ladder is placed againts a wall.If the top of the ladder is sliding down at a rate of 2 feet per second,then estimate the rate at which the bottom of the ladder is moving away from the wall when the bottom of the ladder is 10feet away from the wall.

Let height and base be h and b respectively.All I could understand was that we have to find db/dt,we are given dh/dt ..can somebody help me with this question?
http://www.mathhelpforum.com/math-he...ted-rates.html

3. Let h(t) be the height at time t, b(t) be length of the base of the ladder from the wall at time t.

Now we have

$h(t)^2+b(t)^2=26^2$

$2h(t)\frac{dh}{dt}+2b(t)\frac{db}{dt}=0$

Since you know that the base is 10 feet away, at time t0 so b(t0)=10. And the ladder is sliding down at -2 feet per second $\frac{dh}{dt}=-2$, at time t0.
So what is $\frac{db}{dt}$ at time t0 ?

Can you finish this one now?

4. Ok so we plug in the values for dh/dt as -2 and find out db/dt?

For eg :

2h(0)(-2)+2b(0)db/dt=0
2h(0)(-2)+2(10)db/dt=0
0+20db/dt=0
20 db/dt=0
db/dt=0..??
I know Iam doing this wrong...where am I going wrong?

5. Originally Posted by mathcalculushelp

Ok so we plug in the values for dh/dt as -2 and find out db/dt?

For eg :

2h(0)(-2)+2b(0)db/dt=0
2h(0)(-2)+2(10)db/dt=0
0+20db/dt=0
20 db/dt=0
db/dt=0..??
I know Iam doing this wrong...where am I going wrong?
h(t) and b(t) tell you that h and b are functions of time ... you do not plug in 0 for t.

when b = 10, h = 24