uni maths! (1 Viewer)

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ha...yes, this questions i found in the past paper, its shitting me


a)
Find vectors v, w E R³ that satisfy all the three conditions below simultaneosuly:

v + w = (7,3,5)
v is parallel to (1,2,3)
w is perpendicular to (1,2,3)


b) the function f: r-->r is defined by
f(x) = [sin(pi. x)]/x whenever x=/=0

what values must f(0) take if f is continous at x=0.

my answer was any real value, considered that its an option to be continous...

c) Let a be a fixed real number. Evaluate

lim_x-->a x^3/2 - a^3/2/x - a

and explain its meaning in relation to the function f defined by f(x) = x^3/2


any help is appreciated..thanks!!!!

 

[wogboy]

a)
Find vectors v, w E R3 that satisfy all the three conditions below simultaneosuly:

v + w = (7,3,5)
v is parallel to (1,2,3)
w is perpendicular to (1,2,3)

v + w = (7,3,5) is equivalent to:
v1 + w1 = 7,
v2 + w2 = 3
v3 + w3 = 5

v parallel to 1,2,3 is equivalent to:
v2 = 2v1
v3 = 3v1

w perpendicular to (1,2,3) is equivalent to:
w . (1,2,3) = 0 (dot product)
w1 + 2w2 + 3w3 = 0

So there we have 6 equations & 6 unknowns, which you can find solutions to. (NB: v1, v2, w1 etc are the components of the vectors v and w)

b) the function f: r-->r is defined by
f(x) = [sin(pi. x)]/x whenever x=/=0

what values must f(0) take if f is continous at x=0.

For f to be continouous at x=0, f(0) must equal to lim (x->0) f(x), use L'Hopital's rule to find that limit.

c) Let a be a fixed real number. Evaluate ...

The limit of what you wrote is just a^(3/2) - a^(1/2) - a.

Did you by some chance forget to put brackets around the numerator & denominator? If so, then you can use L'Hopital's Rule to find the limit to be: 3sqrt(a)/2. This happens to be f'(a) (remember the definition of the derivative)

 

[Constip8edSkunk]

a) v = (k,2k,3k), k E R, k=/= 0
w = (7-k, 3-2k, 5-3k)

v . w = 0 since v,w perpendicular
7k - k^2 + 6k - 4k^2 + 15k - 9k^2
= 28k - 14k^2 = 0
k = 2
.'. v = (2,4,6); w = (5, -1, -4)

b) lim{x->0+}f(x) and lim{x->0-}f(x) = -pi (l'hospital)
so a removable discontinuity at x=0. to remove it, let f(0) = limit which is -pi

c)indeterminate for, use l'hospital
= lim{x->a} 1.5sqrt[x] / 1
= 3sqrt[a] / 2

this is the application of the 1st principle of calculus on f(x) and f'(x) = 3f(x)/2


edit: hmmm a bit slow i was...

 

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Originally posted by Constip8edSkunk
7k - k^2 + 6k - 4k^2 + 15k - 9k^2
= 28k - 14k^2 = 0
k = 2
.'. v = (2,4,6); w = (5, -1, -4)

woah..where did u come up with that formula?!?!?!?

and thanks for the help guys

 

[clive]

that's just the result of dotting w and v

ie; (7-k , 3 - 2k , 5 - 3k ) . ( k, 2k, 3k )

= k(7-k) + 2k(3-2k) + 3k(5-3k)

= 28k - 14k^2

 

[...]

hha..err one more =p

let
..........sqrt(x²+x³) / x when x E (0.1]
f(x) = 1 when x = 0
..........a.sqrt(x²+x³) / x when x E [-1,0)

Find a so that the funciton f(x) is continous at x = 0


hmm..btw, how would u approach

lim(x->0) sqrt(x²+x³) / x

 

[wogboy]

hmm..btw, how would u approach ...

sqrt(x^2 + x^3) / x
= sqrt(x^2 * (1 + x)) / x
= (sqrt(x^2) / x ) * sqrt(1 + x)
= |x|/x * sqrt(1 + x)
= signum(x) * sqrt(1 + x)

where signum(x) = 1 for x>0, signum(x) = -1 for x<0, and is undefined for x=0.

taking the right hand side limit:
lim {x->0+} sqrt(x^2 + x^3) / x
= lim {x->0+} signum(x) * sqrt(1+x)
= 1

and taking the left hand side limit:
lim{x->0-} sqrt(x^2 + x^3) / x
= lim {x->0-} signum(x) * sqrt(1+x)
= -1

clearly to make the function continuous, a = -1

since lim {x->0+} f(x) = 1 for all a,
and lim {x->0-} f(x) = -a, and these two limits must be equal for the function to be continuous.

 

[...]

ahhhhhh..

very nice

another vector one..

7. Consider the points A = (1, 3, 2), B = (2, 4, 1),C = (4, 3, 4) and D = (2, 3, 5).
a) Find a normal vector to the plane passing through A,B and C.
(A normal vector to the plane is a vector perpendicular to the plane.)
b) Use part (a) to find an equation for the plane passing through A,B and C.

Solution. a) −! AB = (1, 1,−1) and −! AC = (3, 0, 2), so a normal vector is n = −! AB × −! AC =
(2,−5,−3).
b) 2x − 5y − 3z = −19

now, i can do a) but for b), where did -19 come from

 

[Constip8edSkunk]

using pt normal form the plane should be
n . (x-a) = 0 where a denotes an arbitrary pt on the plane
(1,1,-1) × (3,0,2) (x - (1,3,2)) = 0
so (2, -5, 3).(x - (1,3,2)) = 0
(2, -5, -3) x = (2,-5,-3). (1,3,2)
2x-5y-3z = 2*1 - 5*3 - 3*2
2x-5y-3z = -19

 

[...]

yea fuck, i just figured it out


fark!!!!!

god i am so stoopid