Flex your maths muscles! (2 Viewers)

[bennyhenny]

i get 8+8i. I used arg(z)=(-pi/4)
i.e. z=√2(cos[-pi/4]+isin[-pi/4])
z⁷=(√2)⁷[cis-7pi/4]
= 8√2(cos[pi/4]+isin[pi/4]) (-7pi/4 + 2pi=pi/4 to make within principal arg)
= 8√2(1/√2+[1/√2]i)
= 8+8i

but -1+i is in the second quadrant, so it's arg should be 3pi/4

34. The equation of a conic is given by x^2-y^2=8

the conic is rotated 45 degrees in the negative direction, derive the equation of the conic in new xy plane

 

[STx]

but -1+i is in the second quadrant, so it's arg should be 3pi/4

34. The equation of a conic is given by x^2-y^2=8

the conic is rotated 45 degrees in the negative direction, derive the equation of the conic in new xy plane

oh damn, i see, thx Ben

 

[Riviet]

I found this integral quite interesting:

∫(1+x²) dx
(1+x⁴)

Small hint:

Spoiler

A clever substitution should do the job.

 

[hyparzero]

I found this integral quite interesting:

∫(1+x²) dx
(1+x⁴)

Small hint:

Spoiler

A clever substitution should do the job.

If i use partial fractions, it becomes quite messy, but i get

1/2sqrt(2)*[arctan(1/2sqrt(2)*(2x - sqrt(2)) + arctan(1/2sqrt(2)*(2x + sqrt(2)]

enlighten me on this substitution...

 

[shimmerz_777]

31 § ln(tanx)*(sec^2 x)*(tanx)dx let a= tanx

§ ln(a) *a *(da/dx)*dx
§ a*ln(a) let u=a, du=1, dv=ln(a) v=1/a

1-§1/a da

1-ln(a) +c
1-ln(tanx) +c


is that right? also id really like to know how to get the squareds and otherthings typed, cause i only know the ^2 and its rather annoying to make it easy to read

EDIT:
also for the above question by rivet, a trig substitution looks really tempting but for some reason i cant seem to simplify it... must go over trig stuff!!!

 

[Riviet]

If i use partial fractions, it becomes quite messy, but i get

1/2sqrt(2)*[arctan(1/2sqrt(2)*(2x - sqrt(2)) + arctan(1/2sqrt(2)*(2x + sqrt(2)]

enlighten me on this substitution...

You're on the right track with the arctan, but partial fractions is not necessary.

also for the above question by rivet, a trig substitution looks really tempting but for some reason i cant seem to simplify it... must go over trig stuff!!!

A trig substitution is not needed either.

 

[shimmerz_777]

i cant see it then... by the way im not particularly sure what you mean about the arctan stuff. care to help me out with that?

 

[Riviet]

Try a substitution of u = x - 1/x

 

[STx]

(21) P(cp, c/p) and Q(cq, c/q) lie on hyperbola xy = c². Tangents TP and TQ contact the hyperbola at P and Q. The point M is the midpoint of the chord PQ.
(a) Find the coordinates of T and M.
(b) KMLT is a rectangle whose sides are parallel to the x and y axes. Show that K and L are points on the hyperbola.

in part a) isnt there two points called 'T' which should be called T₁ and T₂,which will turn out to be (2cp,0) and (2cq,0) respectively.

Midpoint_PQ:M: [c(p+q)/2 , c(p+q)/2pq]

 

[Riviet]

in part a) isnt there two points called 'T' which should be called T₁ and T₂,which will turn out to be (2cp,0) and (2cq,0) respectively.

The tangents at P and Q are drawn from T, which implies one external point.

Your midpoint, M is correct.

 

[STx]

The tangents at P and Q are drawn from T, which implies one external point.

Your midpoint, M is correct.

Oh i see, the q wasnt clear enough for me. So equating the tangents at P and Q and solving to get x and y, T[2cpq/p+q , 2c/p+q]