A few problems from Fitzpatrick- care to help? (1 Viewer)

[Candypants]

I recently did Test Paper 1 from Fitzpatrick's book and am stumped by a few of the questions, largely due to the fact that the book only puts solutions, not worked solutions.
If anyone is kind enough to help explain only one or a few of them/ answer them in 'worked-solution' fashion, that would be very much appreciated... if not, I guess that's just my bad luck.

Thanks for your consideration.


3. iii) If cos x = -2/3, and 90dgs < x < 180dgs, write down the exact value of:
a) sin x
b) cot x


7. i) If y'=2y and y=10 when x=0, express y as a function of x.


8. ii) A particle is moving along the X-axis. It started fromt rest at time t=0 from the point x=3. If its acceleration at time t=1+t, find the position pf the particle at time t=2.


9. iii) Solve the quadratic equation (x^2 + 5x)^2 - 84 = 8(x^2 + 5x)


10. i) Show that the locus of a point P(x,y) which moves so that its distance from the line x=8 is twice is distance from the point (2,0) is 3x^2 + 4y^2 = 48


* dgs = degrees; x^2 = x squared

 

[acmilan]

7. y' = 27, by integrating, y = 27x + c, when x = 0, y = 10 so c = 10 hence y = 27x + 10

8.ii) a = 1 + t, by integrating, v = t +(1/2)t^2 + c(1), when t = 0, v = 0 so c(1) = 0.
v = t + (1/2)t^2, by integrating, x = (1/2)t^2 + (1/6)t^3 + c(2), when t = 0, x = 3 so c(2) = 3
x = (1/2)t^2 + (1/6)t^3, when t = 2, x = 2 + (8/6) = 10/3

9. iii) (x^2 + 5x)^2 - 84 = 8(x^3 + 5x)
Let U = x^2 + 5x

hence U^2 - 8U - 84 = 0
U = -6 or U = 14
so x^2 + 5x = -6 OR x^2 + 5x = 14

Solve these two quadratic equations to find the values of x

 

[Candypants]

Thank you SO much for your help; except I made a typo with 7.

As you can see I've changed it now, but it's meant to be y'=2y, and so on.

But I appreciated your help with the other questions and now understand it. So thanks again.

 

[JamiL]

i cant get 7...u confused me n i dont get 8
i realy should start studing... lol
i want that band 6 but im ding 4u next year

 

[Heinz]

3. see diagram

7. dx/dy = 1/2y
x= (1/2)lny + C x = 0, y = 10
:. C = -(1/2)ln10
2x = lny - ln 10
2x + ln10 = lny
y = e^{2x + ln10}
= e^2x. e^ln10
=10e^2x

 

[Candypants]

7. dx/dy = 1/2y
x= (1/2)lny + C

Thanks for your help. Except, I don't get, why does dx/dy = 1/2y?
I don't understand these two lines. I don't think I've ever come across this work before which is why I'm a little slow.

 

[sneaky pete]

"7. i) If y'=2y and y=10 when x=0, express y as a function of x."

hey,
you know how y' is the same as dy/dx ?
just a different notation

we can say:
dy/dx = 2y
we can then invert it to get:
dx/dy = 1/(2y)
(ie flip both sides)

dx/dy = 1/(2y)
times both sides by dy
dx = 1/(2y) dy
integrate both sides:
x = Integral of (1/(2y)) , dy

you know how if the top of the fraction is the derivative of the bottom of the fraction,
the integral is the log of the bottom?
(ie remember the integral of 1/x is log(x) to the base e plus a constant, K)
x = Integral of (1/(2y)) w.r.t y
x = 1/2 Integral of 2/(2y) w.r.t y (we take out a constant of a half, then times the top by 2), the net result is the same as what we started with)
x = 1/2 Integral of 1/(1y) w.r.t y
x = 1/2 log(y) + a constant of integration

y=10 when x=0,
0 = 1/2log(10) + K
so to get them to add up to get 0, K must be equal to Negative (1/2log(10))
x = 1/2 log(y) + a constant of integration
x = 1/2 log(y) + (Negative (1/2log(10)))
x = 1/2 log(y) - 1/2log(10)
times the lot by 2
x = 1/2 log(y) - 1/2log(10)
2x = log(y) - log(10)
2x = log(y/10)

so
e^(2x) = e^(log(y/10))
e to the power of log something, just equals that 'something'
e^(log(y/10)) = y/10

e^(2x) = y/10
times both sides by 10 to get
10e^(2x) = y


i think this is way too difficult for a 2unit exam, i dont think manipulation of the dy and dx signs is even in the 2u syllabus?

 

[Candypants]

Yeh that's what I said I haven't come across manipulation of the dy/dx before. But Fitzpatrick is a hard book - our 3unit people use it for their 2unit course; I just have a copy. Obviously if it was asked it is in the syllabus... but by doing past/test papers, you realise there are a lot of things teachers don't teach you. For instance, I didn't get taught how to apply simpson's/ trapezoidal rule to to Volume.

But thanks a lot for your help GRessie.