Re: rate of change problem

Hi koplan,

For the first question, you will need to use similar triangle to help you solve the answer.

Let's say r is the radius and h is the height. Then you will need to use similar triangle to write r in term of h (like comparing two fractions).

Then you need to derive the volume equation in terms of h.

Then you will probably need to plug in the h that you found when is half volume.

Question 2:

You will need to use sin function to derive.

Since sin t = opp/hyp

t is your angle

opp is the height

hyp is the ladder which is constant

Hope this help :)

Francis

Tutorat A+ Tutoring | Go for A+ , A big step toward the future.

Re: rate of change problem

For the first one draw the cone and label. mark height and corresponding radius also. By similar triangles we get 4h/r = 20/5=4 that gives r = (1/4 ) h

Volume of cone = 1/3 pi r^2 h = 1/3 pi ( 1/4)^2 h^2 h = 1/48 pi h^3

Now differentiate dV/dt = d [1/48 pi h^3] / dt = 1/48 pi 3 h^2 dh / dt. It is given that volume is decreasing by 5 cm^3 / s therefore we have

- 5 = 1/48 pi 3 h^2 dh / dt Now I am sure you can proceed further.

Re: rate of change problem

Thank guys, the 1st one was solved correctly :-)

But I'm still stuck with the 2nd one. With sin function, I have z = sint = h/5, so dz/dh = 1/5, that means (dz/dh).(dh/dt) = (1/5).(-1) = -1/5 (constant), so regardless of the height (h), the rate of change in sint is always -1/5 per sec?. And by the way, how can I deduce the rate of change of the angle from the rate of change of its sin?

Re: rate of change problem

Quote:

Originally Posted by

**kopklan**

A 5m ladder is held against the wall. The top of the ladder is sliding down the wall at the rate of 1m/s. Find the rate af change of the angle between the ladder and the ground when the vertical distance between the top of the ladder and the foot of the wall is 3m.

if the distance from wall is x,

x^2 + h^2 = 25

x = √(25-h^2)

when h=3, x = √(25-9) = √16

sinθ = h/5

2x dx/dt + 2h dh/dt = 0

So, dh/dθ

h/5 = sinθ

h = 5sinθ

dh/dθ = 5cosθ = 5(x/5) = 5(√16/5)= 4

Let me know if that was helpful

edit: No, it wasn't - I did a mistake

Re: rate of change problem

I think your use of "t" is confusing you! On the left side of the equation, sin(t), you mean t to be the angle but when you differentiate, you are using t as the time.

Instead write the angle as " " so that so that .

Re: rate of change problem

with cos(θ).(dθ/dt) = -1/5, then dθ/dt = (-1/5).(1/cosθ). When the vertical distance between top of the ladder and the foot of the wall is 3m, then cosθ = 4/5, so dθ/dt = (-1/5).(5/4) = -1/4. Am I right? if I am, how can I put this identity (dθ/dt = -1/4) into words?

Re: rate of change problem

Re: rate of change problem

Quote:

Originally Posted by

**kopklan** with cos(θ).(dθ/dt) = -1/5, then dθ/dt = (-1/5).(1/cosθ). When the vertical distance between top of the ladder and the foot of the wall is 3m, then cosθ = 4/5, so dθ/dt = (-1/5).(5/4) = -1/4. Am I right? if I am, how can I put this identity (dθ/dt = -1/4) into words?

do you mean "(dθ/dt = -1/4) into words" ? it's the rate of change of the angle with respect to time, and it's negative 1/4 m per seconds because it descends,

hope this helps

Re: rate of change problem

Quote:

Originally Posted by

**dokrbb** do you mean "(dθ/dt = -1/4) into words" ? it's the rate of change of the angle with respect to time, and it's negative 1/4 m per seconds because it descends,

hope this helps

Not quite. Because it is an angle it is -1/4 **radians** per second, not meters.

Re: rate of change problem

Quote:

Originally Posted by

**HallsofIvy** Not quite. Because it is an angle it is -1/4 **radians** per second, not meters.

thanks HallsofIvy, always I have problems with the radians vs all other