Maths problem, max value of a function...
11 January 2009, 04:11 PM
#1
Maths problem, max value of a function...
One of my maths questions I am stuck on is, What is the maximum value of the function....
y= 4 + sin(2t) - 8cos(2t)
Now as far as I understand sin and cos have maximum values of 1 and minimum of -1. The frequency (2t) wont effect the amplitude so I end up with 4+1-(8*1) = -3
This doesnt seem right, any helpers?
Cheers, The Hoff
y= 4 + sin(2t) - 8cos(2t)
Now as far as I understand sin and cos have maximum values of 1 and minimum of -1. The frequency (2t) wont effect the amplitude so I end up with 4+1-(8*1) = -3
This doesnt seem right, any helpers?
Cheers, The Hoff
11 January 2009, 05:17 PM
#2
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Ages ago since I did this, but one method is to simplify the expression so you have only one function expressed in terms of sin.
The maxmium value of -8cos(2t) will be +8 and this will be when 2t=180 degrees. The only question is if moving slightly would allow the sin(2t) to add more than the -8cos(2t) would lose.
So the most elegant way is to simplify the expression so it can be solved.
The maxmium value of -8cos(2t) will be +8 and this will be when 2t=180 degrees. The only question is if moving slightly would allow the sin(2t) to add more than the -8cos(2t) would lose.
So the most elegant way is to simplify the expression so it can be solved.
11 January 2009, 07:29 PM
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When cos(2t) = -1 then sin (2t) = 0
q.e.d.
dunx
P.S. Easy if you plot the function.... start at 90 degree increments and add more for detail, works best for continuous functions.
q.e.d.
dunx
P.S. Easy if you plot the function.... start at 90 degree increments and add more for detail, works best for continuous functions.
Last edited by dunx; 11 January 2009 at 07:31 PM.
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11 January 2009, 08:29 PM
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y= 4 + sin(2t) - 8cos(2t)
Differentiate:
dy/dt = 2 cos 2t + 16 sin 2t
At a maximum or minimum, dy/dt = 0, so:
0 = 2 cos 2t + 16 sin 2t
0 = cos 2t + 8 sin 2t
Double angle formula: sin 2t = 2 sin t cos t
Therefore, 0 = cos 2t + 16 sin t cos t
Another double angle formula: cos 2t = 1 - 2 sin^2 t
So, 0 = 1 - 2 sin^2 t + 16 sin t cos t
2 sin^2 t - 16 sin t cos t = 1
Trig identity sin^2 t + cos^2 t = 1, therefore cos t = +/- SQRT (1 - sin ^2 t)
Therefore,
2 sin^2 t - 16 sin t SQRT (1 - sin^2 t) = 1
Let x = sin t, therefore
2x^2 - 16x SQRT (1 - x^2) = 1
2x^2 - 1 = 16x SQRT (1-x^2)
Square both sides:
4x^4 - 4x^2 + 1 = 256 x^2 (1-x^2) = 256 x^2 - 256 x^4
260 x^4 - 260 x^2 + 1 = 0
Substitute z = x^2,
260z^2 - 260z + 1 = 0
Quadratic formula:
z = (260 +/- SQRT (260^2 - 4x260x1)) / 520
z =0.003861 or 0.99614
Therefore, x = 0.0621 or 0.9981
Therefore, t = 0.0621 or 1.509
Substitute back into the original equation gives a maximum @ t = 1.509 of 12.06.
I think I need a beer
Differentiate:
dy/dt = 2 cos 2t + 16 sin 2t
At a maximum or minimum, dy/dt = 0, so:
0 = 2 cos 2t + 16 sin 2t
0 = cos 2t + 8 sin 2t
Double angle formula: sin 2t = 2 sin t cos t
Therefore, 0 = cos 2t + 16 sin t cos t
Another double angle formula: cos 2t = 1 - 2 sin^2 t
So, 0 = 1 - 2 sin^2 t + 16 sin t cos t
2 sin^2 t - 16 sin t cos t = 1
Trig identity sin^2 t + cos^2 t = 1, therefore cos t = +/- SQRT (1 - sin ^2 t)
Therefore,
2 sin^2 t - 16 sin t SQRT (1 - sin^2 t) = 1
Let x = sin t, therefore
2x^2 - 16x SQRT (1 - x^2) = 1
2x^2 - 1 = 16x SQRT (1-x^2)
Square both sides:
4x^4 - 4x^2 + 1 = 256 x^2 (1-x^2) = 256 x^2 - 256 x^4
260 x^4 - 260 x^2 + 1 = 0
Substitute z = x^2,
260z^2 - 260z + 1 = 0
Quadratic formula:
z = (260 +/- SQRT (260^2 - 4x260x1)) / 520
z =0.003861 or 0.99614
Therefore, x = 0.0621 or 0.9981
Therefore, t = 0.0621 or 1.509
Substitute back into the original equation gives a maximum @ t = 1.509 of 12.06.
I think I need a beer
11 January 2009, 08:38 PM
#11
y= 4 + sin(2t) - 8cos(2t)
Differentiate:
dy/dt = 2 cos 2t + 16 sin 2t
At a maximum or minimum, dy/dt = 0, so:
0 = 2 cos 2t + 16 sin 2t
0 = cos 2t + 8 sin 2t
Double angle formula: sin 2t = 2 sin t cos t
Therefore, 0 = cos 2t + 16 sin t cos t
Another double angle formula: cos 2t = 1 - 2 sin^2 t
So, 0 = 1 - 2 sin^2 t + 16 sin t cos t
2 sin^2 t - 16 sin t cos t = 1
Trig identity sin^2 t + cos^2 t = 1, therefore cos t = +/- SQRT (1 - sin ^2 t)
Therefore,
2 sin^2 t - 16 sin t SQRT (1 - sin^2 t) = 1
Let x = sin t, therefore
2x^2 - 16x SQRT (1 - x^2) = 1
2x^2 - 1 = 16x SQRT (1-x^2)
Square both sides:
4x^4 - 4x^2 + 1 = 256 x^2 (1-x^2) = 256 x^2 - 256 x^4
260 x^4 - 260 x^2 + 1 = 0
Substitute z = x^2,
260z^2 - 260z + 1 = 0
Quadratic formula:
z = (260 +/- SQRT (260^2 - 4x260x1)) / 520
z =0.003861 or 0.99614
Therefore, x = 0.0621 or 0.9981
Therefore, t = 0.0621 or 1.509
Substitute back into the original equation gives a maximum @ t = 1.509 of 12.06.
I think I need a beer
Differentiate:
dy/dt = 2 cos 2t + 16 sin 2t
At a maximum or minimum, dy/dt = 0, so:
0 = 2 cos 2t + 16 sin 2t
0 = cos 2t + 8 sin 2t
Double angle formula: sin 2t = 2 sin t cos t
Therefore, 0 = cos 2t + 16 sin t cos t
Another double angle formula: cos 2t = 1 - 2 sin^2 t
So, 0 = 1 - 2 sin^2 t + 16 sin t cos t
2 sin^2 t - 16 sin t cos t = 1
Trig identity sin^2 t + cos^2 t = 1, therefore cos t = +/- SQRT (1 - sin ^2 t)
Therefore,
2 sin^2 t - 16 sin t SQRT (1 - sin^2 t) = 1
Let x = sin t, therefore
2x^2 - 16x SQRT (1 - x^2) = 1
2x^2 - 1 = 16x SQRT (1-x^2)
Square both sides:
4x^4 - 4x^2 + 1 = 256 x^2 (1-x^2) = 256 x^2 - 256 x^4
260 x^4 - 260 x^2 + 1 = 0
Substitute z = x^2,
260z^2 - 260z + 1 = 0
Quadratic formula:
z = (260 +/- SQRT (260^2 - 4x260x1)) / 520
z =0.003861 or 0.99614
Therefore, x = 0.0621 or 0.9981
Therefore, t = 0.0621 or 1.509
Substitute back into the original equation gives a maximum @ t = 1.509 of 12.06.
I think I need a beer
Now i know how i ended up being a roofer......
11 January 2009, 08:40 PM
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Nice one Andy 
I wish I could remember more of my calculus 
Maybe that's a target for this year At least I knew what needed to be done, just couldn't remember all my trig calculus 
I wish I could remember more of my calculus Maybe that's a target for this year At least I knew what needed to be done, just couldn't remember all my trig calculus
Last edited by Hanslow; 11 January 2009 at 08:42 PM.
11 January 2009, 09:26 PM
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So 2t is about 7 degrees less than my coarse estimate of 180. Interesting.
I like the neat way of doing it, the trouble is you need to have lots of practice and intimate present knowledge of all the trig substitutions to get to your solution. This was the area I used to have most trouble with. It is like trying to solve a chess problem that says "solve for checkmate in 10 moves" in a complex position. My brain just melts.
I like the neat way of doing it, the trouble is you need to have lots of practice and intimate present knowledge of all the trig substitutions to get to your solution. This was the area I used to have most trouble with. It is like trying to solve a chess problem that says "solve for checkmate in 10 moves" in a complex position. My brain just melts.
12 January 2009, 07:30 PM
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I ended up needing a drink...
alcoholic dunx
P.S. Hoff buy a decent bloody Casio, and it will plot it for you !
I use Excel myself....
HTH
dunx
alcoholic dunx
P.S. Hoff buy a decent bloody Casio, and it will plot it for you !
I use Excel myself....
HTH
dunx
Last edited by dunx; 12 January 2009 at 07:32 PM.
12 January 2009, 10:10 PM
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Its sh*t like that, that meant I only got a D in my maths A-level 18 years ago.
I still have yet to find or use pure maths in my work or life since then. Thank ****.
I still have yet to find or use pure maths in my work or life since then. Thank ****.
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