Need help solving problems (1 Viewer)

Bokky

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hi guys, just skimming thru my 3U half yearly which i did pretty poorly in, ive got a few questions from it which will appear in my next test so i gotta learn them now.

1) parabola x^2 = 4y. The tangent to the parabola at P(2p, p^2), p>0 cuts the x axis at A. The normal to the parabola at P cuts the y axis at B.

(i) Derive the equation of the tangent AP.

(ii) Show that B has the coordinates (0, p^2+2).

(iii) Let C be the midpoint of AB. Find the cartesian equation of the locus of C.


2) The region bounded by y = 3sin x, the x axis and the line x= pie/2 is rotated about the x axis to form a solid. Calculate the volume of solid (leave answer in exact form).

3)The following area is rotated about the y-axis.

y=3x^2 ; y=16-x^2

Show that the volume of the solid of revolution is 32pie units^3

4) (i) By equating the coefficients of sinx and cosx or otherwise, find the constants A and B satisfying the identity
A(2sinx+cosx)+B(2cosx-sinx)==sinx+8cosx

(ii) Hence evaluation int. (sinx+8cosx)/(2sinx+cosx)dx

LOL have fun, ta guys :D
 

FinalFantasy

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1) parabola x^2 = 4y. The tangent to the parabola at P(2p, p^2), p>0 cuts the x axis at A. The normal to the parabola at P cuts the y axis at B.

(i) Derive the equation of the tangent AP.

(ii) Show that B has the coordinates (0, p^2+2).

(iii) Let C be the midpoint of AB. Find the cartesian equation of the locus of C.

y=x²\4
y'=x\2
i)eq. of tangent at P:
(y-p²)\(x-2p)=(2p)\2=p
y-p²=px-2p²
.: y=px-p²

ii)eq. of normal at P:
(y-p²)\(x-2p)=-1\p
py-p³=-x+2p
x+py-p³-2p=0
since this line cuts y axis at B, the x coordinate should be 0 at the Y axis.
sub x=0 in, u get
0+py-p³-2p=0
y=p²+2
.: B=(0, p^2+2)
iii)to get mid point AB u need coordinates of A and B.
for coords of A, the tangent y=px-p² cuts x-axis at A
so y coordinate at x-axis is 0
i.e: 0=px-p²
x=p
.: A=(p, 0)
MidPT of AB:
x=p\2 and y=(p²+2)\2
p=2x
.: y=(4x²+2)\2 is da eq. of locus C
 

Trev

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Bokky said:
1) parabola x^2 = 4y. The tangent to the parabola at P(2p, p^2), p>0 cuts the x axis at A. The normal to the parabola at P cuts the y axis at B.

(i) Derive the equation of the tangent AP.

(ii) Show that B has the coordinates (0, p^2+2).

(iii) Let C be the midpoint of AB. Find the cartesian equation of the locus of C.
(i)
x² = 4y
y = x²/4
dy/dx = x/2
Gradient at point P(2p,p²) = 2p/2 = p.
Equation tangent:
(y-p²)=p(x-2p)
y = px - p²

(ii)
Gradient normal = -1/p.
Equation normal:
(y-p²)=-1/p(x-2p)
x + py = p<sup>3</sup> + 2p

B where x = 0.
py = p<sup>3</sup> + 2p
y = p² + 2.
&there4; B(0,p²+2).

(iii)
B (0,p²+2).
A (p,0).
Midpoint AB (p/2, [p²+2]/2) is C.
Cartesian equation: (I think this is what a cartesian equation is :p)
2x=p
2y = p²+2
&there4; 2y = (2x)² + 2
y = 2x² + 1.
 

FinalFantasy

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2) The region bounded by y = 3sin x, the x axis and the line x= pie/2 is rotated about the x axis to form a solid. Calculate the volume of solid (leave answer in exact form).

V=pi int. (3sinx)² dx from x=0 to x=pi\2
V=9pi int. sin²x dx=9\2pi int. (1-cos2x) dx
=9pi\2 [x-(1\2)sin2x] from 0 to pi\2
juz put da terminals in den ur done
 

FinalFantasy

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3)The following area is rotated about the y-axis.

y=3x^2 ; y=16-x^2

Show that the volume of the solid of revolution is 32pie units^3

curves intersect (-2, 12) and (2,12)
y\3=x²
x²=16-y

V=pi int. (y\3) dy from 0 to 12 + pi int. (16-y) dy from 12 to 16
=pi [y²\6] 0-12 + pi[16y-y²\2] 12-16
=24pi+pi(128-120)=24pi+8pi=32pi
 

FinalFantasy

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4) (i) By equating the coefficients of sinx and cosx or otherwise, find the constants A and B satisfying the identity
A(2sinx+cosx)+B(2cosx-sinx)==sinx+8cosx

(ii) Hence evaluation int. (sinx+8cosx)/(2sinx+cosx)dx


A(2sinx+cosx)+B(2cosx-sinx)==sinx+8cosx
2Asinx+Acosx+2Bcosx-Bsinx==sinx+8cosx
sinx(2A-B)+cosx(2B+A)=sinx+8cosx
equating coefficients of like powers:

2A-B=1 -----> 4A-2B=2
2B+A=8
4a+A=2+8
5A=10
.: A=2
2B+2=8
2B=6
.: B=3
Now
2(2sinx+cosx)+3(2cosx-sinx)=sinx+8cosx
(ii) Hence evaluation int. (sinx+8cosx)/(2sinx+cosx)dx
=int. [2(2sinx+cosx)\(2sinx+cosx)+3(2cosx-sinx)(2sinx+cosx)] dx
=int. 2 dx+3int. (2cosx-sinx)(2sinx+cosx) dx
=2x+3 ln |2sinx+cosx|+C
 

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