S.H.M - Netwon's Law of Cooling - Inverse Functions (1 Viewer)

[nick1048]

If anyone could help me complete these questions to tie up the ends of my work that would be fantastic!!!

Simple Harmonic Motion

1) Assume that the tides rise and fall in S.H.M. A ship needs 10 metres of water to pass down a channel safely. At low tide the channel is 9m deep and at high tide 12m deep. Low tide is at 9 a.m. and high tide at 4 p.m. At what time can the ship proceed safely?

ANSWER: Between 11.44 a.m. and 8.16 p.m.

2) Assume that over several days of constant weather the cycle of temperatures each day is S.H between 13 degrees at 4 a.m. and 23 degrees at 4 p.m. At what times of the day would the temperature be:
a) 18 degrees
b) 15 degrees
c) 21 degrees

ANSWER:
a) 10 a.m.
b) 7.33 a.m.
c) 12.27 p.m.

Newton's Law of Cooling

1) A body, initially at room temperature 20 degrees Celcius, is heated so that its temperature would rise by 5 degrees Celcius/min if no cooling took place. Cooling does occur in accordance with Newton's Law of Cooling and the maximum temperature the body could attain is 120 degrees Celcius. How long would it take to reach a temperature of 100 degrees Celcius?

ANSWER: 32.2 min

Inverse Functions

1) f(x) = x^2 + 2x , x>= 0

ANSWER: Root (x + 1) - 1, x >= 0, f^-1 (x) >= 0

Now I know the last one is simple but I just can't get my head around it. Thanks alot guys! Your help is seriously appreciated!!!

nick1048

 

[nick1048]

ok maybe i should have included the answers... I'll edit the thread with the correct answers

 

[maths > english]

solution to the first one is attached below

 

[FinalFantasy]

Newton's Law of Cooling

1) A body, initially at room temperature 20 degrees Celcius, is heated so that its temperature would rise by 5 degrees Celcius/min if no cooling took place. Cooling does occur in accordance with Newton's Law of Cooling and the maximum temperature the body could attain is 120 degrees Celcius. How long would it take to reach a temperature of 100 degrees Celcius?
dT\dt=5-k(T-20)
dT\dt=0 when T=120
.: 0=5-k(100)
100k=5
k=1\20

dT\dt=5-1\20(T-20)=5+1-T\20
=6-T\20
=-1\20(T-120)
.: T=120+Ae^(-1\20)t
when t=0, T=20
.: A=-100
.: T=120-100e^(-1\20)t
when T=100, 100=120-100e^(-1\20)t
-20=-100e^(-1\20)t
ln (1\5)=(-1\20)t
.: t=20 ln (5)

Inverse Functions

1) f(x) = x^2 + 2x , x>= 0

inverse:
x=y²+2y
y²+2y-x=0
y=-2+-sqrt(4+4x) all over 2
=-2+-2sqrt(1+x) all over 2
=-1+-sqrt(1+x)
den u reason to take the positive so f^-1 (x)=sqrt(1+x)-1

 

[nick1048]

champions, I'll check this out in the morning lol I'm so tired of maths today

 

[fahadmumtaz88]

answer to inverse function

the answer to that last question is

f(x) = x^2 + 2x
which means y= x^2 + 2x

Interchanging x with y,you get

x = y^2 + 2y.

and now make y the subject of the formula, using completing the squares
therefore,

x + 1 = y^2 + 2y +1

x + 1 = (y + 1)^2

therefore y= sqrt(x + 1) -1

 

[maths > english]

sorry, forgot to answer the second SHM question, answer attached

 

[nick1048]

woah those shm are way too long for exam format... my friend had an alternative approach, thanks guys!

 

[nick1048]

ok maths > english... thanks for ur help but your not answering the question. Your only giving one part of the answer... and I don't really understand where your putting stuff and why... The SHM Questions remain unanswered and the worst part is, if you look at the 2004 HSC for 3u maths the questions are similar to these ones -_-' Im doomed.

 

[maths > english]

because the minimum value is 9 and the maximum 12 and the object moves with SHM, the object is oscillating about the line x=10.5 because 10.5 is in the middle of 9 and 12

the amplitude is 1.5 because 9 = 10.5 - 1.5 and 12 = 10.5 + 1.5 (1/2 the difference between the max and min values)

half of a full wavelength is inbetween the max and min values so half the period is the time from 9am to 4pm (7hrs)

therefore period = 14hrs = 2*pi/w, w = pi/7

therefore the equation is of the form 1.5 sin (pi/7*t + @) + 10.5

the rest is in the attached images above and the second question applies the same techniques

i hope that helps, good luck!

 

[nick1048]

ok I pretty much understand everything except the last bit involving the alpha angle...

omg wrong substitution... I gets it now. One question, have you had a look at the 04 3u HSC paper... omg man. The sections that take care of SHM and Projectile Motion are incredibly difficult... Im doomed...

 

[FinalFantasy]

ok I pretty much understand everything except the last bit involving the alpha angle...

omg wrong substitution... I gets it now. One question, have you had a look at the 04 3u HSC paper... omg man. The sections that take care of SHM and Projectile Motion are incredibly difficult... Im doomed...

dun worry man, just do a few more exercises on them den they'll become easy to u..

 

[nick1048]

lol thanks, I actually got all of the questions out btw... your right, practice makes familiar aspects flood back. I should be orite for my exam tomoro now, as you can see, maths isn't exactly my strongest point :uhhuh: lol.