H2 maths question(rates of change)

lavastar SG

1)A train, starting at 1100 hours travels east at 75km/h while another , starting at noon from the same point, travels south at 100km/h. How fast are they seperating in 1500 hours?

2) A circular cylinder is expanding in such a way that at time t seconds, the length of the cylinder is 20x cm3 and the area of the cross section is x cm2. Given that when x=5, area of cross section is increasing at a rate 0.025 cm2/s. Find the rate of increase at this instant of the

a)length of cylinder
b)volume of cylinder
c)radius of cylinder

This question im unsure of what the length means( diameter or height ) .

Any help please? thx

eagle SG

Can we have the answers next time if you have them when you post the questions?

lavastar SG

i have only the answer for the first part : 87.5 X sq root 2 ..

sry should have included answers for u all Smile

tanjun SG

1st qn.
lol. I can't get the ans. Anyway, this is how i can do. Any errors?

The first train travels east and the second train travels south. Imagine the first train as travelling to the right if North is in upward direction and so the second train will be travelling downwards.
Thus, it will look like a right angle triangle.

Speed at which they are separating = square root( 75^2 + 100^2)
= 125 km/h

eagle SG

I a bit busy to try... but do note that... one started at 1100 and the other started at 1200

tanjun SG

dun really understand. Does the qn mean that they wan the speed of separation at 1500?

eagle SG

means at 1500, they are both 300km from the starting point

Dunno if can see my working... A bit blur:

Right click on image and select view image to see the whole image

tanjun SG

Your unit is wierd for your length of cylinder. Is it cm^3 or cm?

eagle SG

take it as cm and try...

me back to article reading

^tamago^ SG

Given: L = 20x cm, A = x cm²
---> L = 20A
---> dL/dA = 20 cm-¹

---> V = AL = 20A² = 20x² cm³
---> dV/dA = 40A cm

A = ½pi*r² cm²
dA/dr = pi*r cm = 2A/r cm

Given: when x=5, dA/dt = 0.025 cm²/s.
---> A = 5 cm²
---> r = sqrt(10/pi)
---> dA/dr = sqrt(10pi)

a) dL/dt
= dL/dA × dA/dt
= 20 cm-¹ × 0.025 cm²/s
= 0.5 cm/s

b) dV/dt
= dV/dA × dA/dt
= 40A × 0.025 cm²/s
= 40(5)cm × 0.025 cm²/s
= 5 cm³/s

c) dr/dt
= dr/dA × dA/dt
= sqrt(10pi) × 0.025 cm²/s
= 0.140 cm/s (3 s.f.)

^tamago^ SG

Let the train departing at 1100hrs be Train A, train departing at noon be Train B, s be the absolute displacement from the same origins at time t and x be the absolute seperation of the 2 trains at time t.

Given:
vA = 75km/h, east
vB = 100km/h, south

when t=4h, sA = vA × tA = 75km/h × 4h = 300km
when t=3h, sB = vB × tB = 100km/h × 3h = 300km

Since the 2 vectors of vA and vB are 90 degrees apart, by Pythagoras' theorem,
x² = (sA)² + (sB)² = 1.8 × 10^5 km²
x = 424km (3 s.f.)

eagle SG

huh... what doing you... How fast are they seperating in 1500 hours become their distance apart?

Btw.. I answered the question in a few lines just above, in that blur blur picture Wink

tanjun SG

Yup. Didn’t know need to use the idea of dx and dy over dt to find the speed of separation during 1500. Well done eagle.

lavastar SG

thanks eagle for the ans Smile differentiation in h2 maths is so hard Mad

^tamago^ SG

oh. "how fast". Laughing

eagle SG

ok la... I didn't expect many to see it... Last time I had the same problems too... This time... I took a while because I couldn't do it mentally, which I do usually in solving the questions here... Only when I draw out the diagrams, did I see the solution.

Moral of the story: Do draw diagrams as often as possible (BIG diagrams)