Differentiation-rate of chang

1
dV/dt=-5
V=(1/3)πr^2h
Consider the triangle formed in the cone
r/h=10/15
r=(2/3)h and (3/2)r
So d/dt((1/3)πr^2h)=-5
d/dt((1/3)π(4/9)h^3)=-5
(4π/27)(3h^2)dh/dt=-5
sub h=10, we have dh/dt=-9 / 80π
(b)
This time
So d/dt((1/3)πr^2h)=-5
d/dt((1/3)π(r^2)(3/2)r)=-5
(π/2)(3r^2)dr/dt=-5
sub r=20/3, we have dr/dt=-3 / 40π
2
(a)
Let the horizontal length is x and the vertical length is y
then x^2+y^2=2.5
2x(dx/dt)+2y(dy/dt)=0
sub x=2,y=1.5,dx/dt=0.4
We have dy/dt=-8/15
So the rate at which the top of the ladder is moving downward is 8/15 m/s
(b)
The mid-point is (x/2,y/2)
The square of distance between the middle of the ladder and the foot of the wall
l=(x/2)^2+(y/2)^2
dldt
=(x/2)dx/dt+(y/2)dy/dt
sub x=2,y=1.5,dx/dt=0.4,dy/dt=-8/15
we have dl/dt=0
and so the result is 0 m/s

(1a) Let h be the water depth, then by similar triangle, the base radius of the water can be found by:

h/r = 15/10
r = 2h/3

So, in terms of h only, volume of water can be found as:

V = πhr^2 / 3
= 4πh^3 / 27

dV/dt = 4πh^2 / 9 dh/dt
-5 = 4πh^2 / 9 dh/dt
dh/dt = -45/(4πh^2)

So when h = 10:
dh/dt = -45/(4π x 10^2)
= -45/400π
= -9/80π cm/s

(b) r = 2h/3
dr/dt = 2/3 (dh/dt)
= -3/40π cm/s

(2a) Let h be the vertical height of the top of the ladder from the ground, then

x = √(2.5^2 - h^2) where x is the horizontal distance of the foot from the wall

dx/dt = 1/[2√(6.25 - h^2)] x [-2h (dh/dt)]
0.4 = -h(dh/dt)/[√(6.25 - h^2)]
dh/dt = -0.4√(6.25 - h^2) / h

When x = 2, h = 1.5 and therefore

dh/dt = -0.4√(6.25 - 1.5^2) / 1.5
= -0.4 (2) / 1.5
= -8/15 m/s

So the top is slipping down at 8/15 m/s.

(b) Since the wall is perp. to the ground, the wall, the ground and the ladder form a right-angled triangle with the ladder being the hypothenus side.

Hence, with the ladder being the diameter of a circle, the foot of the wall always lies on this circle as for reason of angle in semi-circle = 90 deg.

Then then middle of the ladder will be the centre of the circle which is supposed to have a fixed distance from the foot of the wall.

Finally, the rate of change of the distance between the middle of the ladder and the foot of the wall is always 0 m/s.