rates of change (1 Viewer)

[shkspeare]

bleh someone help me =(

1. A 4m long ladder is leaning against a wall. Its base is slipping away from the wall at a constant rate of 2m/s. At what rate, to 2 decimal places, will the top of the ladder by slipping down the wall when the base is 1m out from the wall?

2. An observer sees a plane directly overhead, 2km above the ground. The plane is moving horizontally at a constant speed of 500km/h. Find the rate at which the distance will be increasing between the observer and the plane when the distance between them is 5km.

3. A kite, 20m high, is blown along horizontally at 3m/s by the wind. At what rate is the string being let out when it is 40m long?

4. A spherical bath oil capsule dissolves in the bath so that its decrease in volume is proportional to its surface area. If its shape remains spherical as it dissolves, show that the radius of the capsule will decrease at a constant rate.

5. A chute drops sand at a cosntant rate of 8m cubed per minute. As the sand falls to the ground it forms a cone shape such that the height of the cone is twice its radius. Find the rate at which the height of the sand will be changing, to 2 decimal places, when its height is 2n.

I have a test coming in a week and im still stuck on these questions ='(

Thx!

 

[Estel]

1.
Firstly, h=rt(16-x^2) where h is the height and x is the width.
dh/dt = dh/dx * dx/dt
= -x/rt(16-x^2) * 2
= -2x/rt(16-x^2)
Now x=1
hence dh/dt=-0.516...
The wall is slipping down at 0.52m/s.

 

[Zarathustra]

Originally posted by Estel
The wall is slipping down at 0.52m/s.

What sort of wall are you thinking of? How do you know (and are willing to attempt) the year 12 course - aren't you in prelim?

 

[CrashOveride]

accelerant obviously

 

[Estel]

I meant ladder lol...
Na I'm not an accelerant.
Just someone trying questions that I probably can't do.
Rates of change is really only juggling differentiation, at least for the q's you guys have been posting.

Should prolly add a disclaimer to every post i do...

 

[zelda]

1) if u draw a diagram up, u kno that ur dealing with a pythagoras problem....so
x^2+y^2=4^2
y^2=4^2-x^2
use implicit differentiation
2y dy/dt = -2x dx/dt
-y/x dy/dt = dx/dt
-root15/1 dy/dt = 2
dy/dt = -2/root15
=-0.52
therefore its moving down at rate of 0.52m/s

2)basically use the same idea and u will get it
3)same applies to this question
x^2+y^2=20^2
y^2=20^2-x^2
2y dy/dt = -2x dx/dt
-2y/2x dy/dt = dx/dt
- y/x dy/dt = 3
-44.7/40 = 3 (by finding hypotenus y...so root2000 u get 44.7)
dy/dt = -3/44.7/40
=2.6m/s

5) volume of cone: V= 1/3pier^2h
since height of cone is twice its radius
then h = 2r, r=h/2
so sub r into V...
u get 1/12pieh^3
differentiate it...dv/dh = 1/4pieh^2

so u have: dv/dt = dv/dh x dh/dt
8 = dv/dh x dh/dt
when h 2
u find dh/dt

and u get 2.55m/min

 

[shkspeare]

thx to all

anyone do 3 or 4?

 

[Xayma]

Originally posted by shkspeare
4. A spherical bath oil capsule dissolves in the bath so that its decrease in volume is proportional to its surface area. If its shape remains spherical as it dissolves, show that the radius of the capsule will decrease at a constant rate.

dV/dt=kS where S is the surface area and k is a constant.

dr/dt=dV/dt*dr/dV

V=(4/3)πr³
dr/dV=1/(dV/dr)
dV/dr=4πr²

dr/dV=1/4πr²
But
dV/dr=S

.:
dr/dt=kS*1/S
=k
and k is a constant (from above)

 

[pc_wizz]

Originally posted by shkspeare
3. A kite, 20m high, is blown along horizontally at 3m/s by the wind. At what rate is the string being let out when it is 40m long?

this is from Maths In Focus Extension One HSC Course

Ex 6.2 q(23)

heres my solution:

D^2 = (x^2) + (y^2) [Pythagoras]
D = 40
y = 20
x^2 = 1200
:. x = sqrt(1200)

dx/dt = 3

D^2 = (x^2) + 400
:. D = sqrt[(x^2) + 400]

dD/dx = (1/2).{[(x^2) + 400]^(-1/2)}.(2x)

dD/dt = dD/dx . dx/dt
= (1/2).{[(x^2) + 400]^(-1/2)}.(2x) . dx/dt
=(3x).{[(x^2) + 400]^(-1/2)}

sub x = sqrt(1200) into equation ...

:. dD/dt = {3[sqrt(1200)]}/40
= 2.60 m/s

pcwizz