Differentiation

Masterith828

2010-01-08 9:07 am

3. An object moves along a horizontal line. Its displacement s from a fixed point at time t is given by s = t^3 -9t^2 +24t. What is the total distance travelled from t=0 to t=7 ?
4. John observes a helicopter is 260m from him. The helicopter is flying horizontally towards him at a rate 30m/s at an attitude of 100m. How fast is the helicopter approaching him ?
5. A tennis player 1.8m tall dashes at a rate of 4m/s towards a flood light post which is 5m high. Find the rate of change of the displacement of the tip of his shadow.
6. A filter is in the form of an inverted cone of base radius 10cm and height 15cm. Water is running out of it at a constant rate of5cm^3/s. When the depth of water is 10cm , find the rate of (a)the water level, (b)the radius of water surface.
7. A ladder 2.5m long leans against a wall. Its foot is 2m from the wall and slipping away at a rate 0.4m/s. Find the rate of change of the distance between the middle of the ladder and the foot of the wall.
8. In a railway station, a train, starting at noon, travels east at 75km/h. Another train, starting at 1 p.m., travels north at 80km/h. How fast are they separating at 2 p.m. ?

回答 (1)

用戶9112

2010-01-09 7:39 am

✔ 最佳答案

(3) s = t^3 - 9t^2 + 24t
ds/dt = 3t^2 - 18t + 24 = 3(t^2 - 6t + 8) = 3(t - 2)(t - 4)
It can be seen that
for 0 < t < 2, ds/dt is positive
for 2 < t < 4, ds/dt is negative
for t > 4, ds/dt is positive
The displacement from 0 <= t <= 2 is 8 - 36 + 48 - 0 = 20
The displacement from 2 <= t <= 4 is (64 - 144 + 96) - 20 = -4
The displacement from 4 <= t <= 7 is (343 - 441 + 168) - 16 = 54
Although the overall displacement is 70, the total distance travelled = 20 + 4 + 54 = 78
(4) Let the horizontal distance of the helicopter from John be x, and actual distance be y
Pythagora's theorem, y^2 = x^2 + 100^2
2y(dy/dt) = 2x(dx/dt)
dy/dt = (x/y)(dx/dt)
when y = 260, 260^2 = x^2 + 100^2 => x = 240
Therefore dy/dt = (240/260)(30) = 27.69 m/s
(5) Let the distance of the man from the foot of the post be y m
Let the distance of the tip of the man's shadow from the foot of the post be x m
(5 - 1.8)/y = 5/x
3.2x = 5y
3.2(dx.dt) = 5(dy/dt)
3.2(dx/dt) = 5(-4)
dx/dt = -6.25 m/s
(6) Let the radius be r and height be h
r/h = 10/15 = 2/3
r = 2h/3
Volume of water V = (1/3)πr^2h
= (1/3)π(2h/3)^2h
= 4πh^3/27
dV/dt = (4πh^2/9)(dh/dt)
dh/dt = (-5)/(4π*100/9) = -0.0358 cm/s
dr/dt = (2/3)(dh/dt) = -0.0239 cm/s
(7) Please check the question, the centre of the ladder is always 1.25m from the foot of the wall.
(8) Let t be the number of hours after noon
The distance between the train D = √[(75t)^2 + (80t - 80)^2]
D^2 = 5625t^2 + 6400t^2 - 12800t + 6400
D^2 = 12025t^2 - 12800t + 6400
2D(dD/dt) = 24050t - 12800
dD/dt = (24050t - 12800)/2D
at t = 2, D = √(48100 - 25600 + 6400) = 170
dD/dt = (48100 - 12800)/(2*170) = 103.8 km/h

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