Rates of Change- cambridge development (1 Viewer)

[jellybelly59]

need help with book 2 exercise 7F question 8b

 

[AlexJB]

Mind typing up the question?

 

[jellybelly59]

Given that the dam is initally empty find W

 

[alakazimmy]

W = 1.2t - cos²(pi/12)*t²/2 + C

When t=0, W=0, hence C=0

Therefore,
W = 1.2t - cos²(pi/12)*t²/2

 

[jellybelly59]

W = 1.2t - cos<SUP>2</SUP>(pi/12)*t<SUP>2</SUP>/2 + C

When t=0, W=0, hence C=0

Therefore,
W = 1.2t - cos<SUP>2</SUP>(pi/12)*t<SUP>2</SUP>/2

sorry not the right answer

 

[AlexJB]

I got the same as alakazimmy.

 

[x.Exhaust.x]

sorry not the right answer

David what's the answer? I also got the same answer as Alakazimmy yesterday (when I told you to integrate it). That's odd...

 

[jellybelly59]

 

[Timothy.Siu]

dw/dt=1.2-cos^2 (txpi/12)
cos 2x=2cos^2 x-1
cos^2 x=1/2 (cos 2x+1)
=1.2-1/2(cos pi/6+1)
=1.2-0.5cos pi/6-0.5

integrating that,
w=0.7t-3/pi x sin pi/6 +C
when t=0 w=0
therefore C=0
therefore w=0.7t-3/pi x sin pi/6

i dont know what the other guys were doing....i guess they forgot/dont know how to integrate trig

 

[alakazimmy]

dw/dt=1.2-cos^2 (txpi/12)
cos 2x=2cos^2 x-1
cos^2 x=1/2 (cos 2x+1)
=1.2-1/2(cos pi/6+1)
=1.2-0.5cos pi/6-0.5

integrating that,
w=0.7t-3/pi x sin pi/6 +C
when t=0 w=0
therefore C=0
therefore w=0.7t-3/pi x sin pi/6

i dont know what the other guys were doing....i guess they forgot/dont know how to integrate trig

Can't blame us if the guy can't get his notation correct.

His notation implies that the variable t is outside of the cosine function.

 

[AlexJB]

So wait, it's cos^2 (t.pi/12)?
The way he's done it it looks like its cos^2 (pi/12)t, which would act as a constant.

 

[alakazimmy]

So wait, it's cos^2 (t.pi/12)?
The way he's done it it looks like its cos^2 (pi/12)t, which would act as a constant.

yup, that's what i just said ^_^

 

[jellybelly59]

lol im sorry about that but it seriously looks the same as what the book looks like and thanks for the help.