- Solve the following system of equations.
x^2 + 4y^2 = 20
xy = 4 - A vector v has a magnitude of 200m and an angle of elevation of 50o. To the nearest tenth of a meter, find the magnitudes of the horizontal and vertical components of v.
- Find the area bounded by the region y = x^2 + 1 , y = -x + 3, x = 0 and x = 3
- Find the derivative of y = x^2 + 1/x
- Find the derivative of y = 1 + 4x^3
- Find the derivative of y = x^2 + 54/x
- Displacement is given by s = 2 + 3t - t^2, find the maximum displacement
- Find the coordinates of the stationary point of y = x^3 -12x - 12
- i) Factorise by trial and error the expression f(x) = 2x^2 + x + 15 ii) Write in the form f(x) = a(x + b)^2 + c iii) Hence find the turning point of f(x) iv) Differentiate 2x^2 + x + 15 hence determine the turning point. v)Comment on iii) and iv) v1) Sketch the curve 2x^2 + x + 15 and determine the factors of the curve and the turning point vii) Using te graph solve 2x^2 + x + 15 = 4
- The curve y^2 = 12x intersects the line 3y = 4x + 6 at 2 points. Find the distance between the 2 points.
Tuesday, November 24, 2009
Last class questions
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let
ReplyDeletex^2 + 4y^2 = 20....(1)
xy = 4...(2)
from 2
x=4/y..(3)
sub 3 into 1
(4/y)^2+4y^2=20
16/y^2+4y2=20
16/y^2=20-4y^2
16=y^2(20-4y^2)
16=20y^2-4y^4
4y^4-20y^2+16=0
/4
y^4-5Y^2+4=0
Factorize
(y^2+4)(y^2-1)
Y^2=-4 or y^2=1
y=sqrt(-4) or y= sqrt 1
does not exist or +/- 1
therfore sub y=1 into (2)
xy=4
1x=4
x=4
or y=-1
xy=4
-1x=4
x=-4
when y=1 x=4
y=-1 x=-4
4.
ReplyDeleteFind the derivative of y = x^2 + 1/x
y=2x2-1+1x-1
y=2x+1
5.
ReplyDeleteFind the derivative of y = 1 + 4x^3
y=1-1+ 4(3)x3-1
y=0+12x2
y=12x^2
6.
ReplyDeleteFind the derivative of y = x^2 + 54/x
y=2x2-1+54x-1
y=2x+54
7.
ReplyDeleteDisplacement is given by s = 2 + 3t - t^2, find the maximum displacement
for max displacement ds/dt=0
so
s=2+3t-t^2
ds/dt=3-2t
therdore 3-2t=0
-2t=-3
2t=3
t=3/2
therfor max displacement is when t=3/2
s=2+3(3/2)-(3/2)^2
s=2+9/2-9/4
s=13/2-9/4
s=17/4
s=4.25m
8.
ReplyDeletey = x^3 -12x - 12
cordinates of stationary pt
dy/dx=3x2-12
3x^2-12=0 ( at st pt =o)
3x^2=12
x^2=4
x=sqrt4
x=2
at x=2 y=2^3 -12(2) - 12
y=8-24-12
-28
x=2 y=-28
ques 10
ReplyDeletey^2 = 12x (1)
3y = 4x + 6 (2)
from eqn 2
y = (4x + 6)/3 (3)
substitute eqn 3 in eqn 1
( (4x/3) + 2)( (4x/3) + 2) = 12x
(16x^2/9) + (16x/3) + 4 = 12x
multiply by 9 throughout
16x^2 + 48x + 36 = 108x
/ by 4
4x^2 + 12x - 27x + 9 = 0
4x^2 - 15x + 9 = 0
4x^2 - 12x - 3x + 9 = 0
4x(x - 3) -3(x - 3) = 0
(x - 3) (4x - 3) = 0
x = 3 or x = 3/4
sub x = 3 in eqn 3
y = (4 * 3 + 6 )/3
6
sub x = 3/4 in eqn 3
y = ( 4* 3/4 +6 )/3
3
points of intersection ( 3 , 6) and ( 3/4 , 3)
distance between points =
( (6 - 3 )^2 - ( 3 - 3/4 )^2 )^0.5
3.75
ques 9 i
ReplyDelete2x^2 + x + 15
(2x + )( x + )
cannot be factorised
9 ii
ReplyDeletef(x) = 2x^2 + x + 15
2( x^2 + x/2 + (1/4)^2 ) + 15 - 2(1/4)^2
2( x + 0.25 )^2 + 14.875
9 iii
ReplyDeletef(x) = 2(x + 0.25 )^2 + 14.875
Turning point (-0.25 , 14.875)
9 iv
ReplyDeletey = 2x^2 + x + 15 (1)
dy/dx = 4x + 1
at turning point dy/dx = 0
4x + 1 = 0
x = -0.25
sub x = -0.25 in eqn 1
y = 2*(-0.25)^2 - 0.25 + 15
14.875
turning point ( -0.25,14.875)
9 v
ReplyDeleteYou can just look at the completed square in iii and determine the turning point but in the differential you have to equate dy/dx to zero and then sub the x value in the eqn with y . It is much more longer.
Question on vectors (for anyone up to the challenge)
ReplyDeleteIf AB is a vector 4i + 10j and AD is a vector 11i - 12j what is their resultant and what would the resultant vector be called ?
To the person who answered my question on vectors could you explain how you arrived at your answer.
ReplyDeletei dont understand what GL do in Q1 from sub come down
ReplyDeleteWell 260....previously from equation 2, GL found the value of x by rearranging the formular to be 4/y. And then, he substituted it into equation 1. in other words, wherever you see x in equation 1, replace it with 4/y and solve.
ReplyDeleteyea fadda thanks for clearin that up for me,but how did he reach to 4y^4-20y^2+16=0
ReplyDeletex^2 + 4y^2 = 20....(1)
ReplyDeletexy = 4...(2)
from 2
x=4/y..(3)
sub 3 into 1
(4/y)^2+4y^2=20
16/y^2+4y2=20
16/y^2=20-4y^2
16=y^2(20-4y^2)
16=20y^2-4y^4
4y^4-20y^2+16=0
/4
y^4-5Y^2+4=0
Factorize
(y^2+4)(y^2-1)
Y^2=-4 or y^2=1
y=sqrt(-4) or y= sqrt 1
does not exist or +/- 1
therfore sub y=1 into (2)
xy=4
1x=4
x=4
or y=-1
xy=4
-1x=4
x=-4
therefore y=1, 4 and x=-1, -4
thankz fadda but i figure it out
ReplyDelete4. y=2x+1
ReplyDelete5. y=12x^2
ReplyDeleteOK.. cool...
ReplyDelete6. y=2x+54
ReplyDeletemax displacement ds/dt=0
ReplyDeletes=2+3t-t^2
ds/dt=3-2t
therfore 3-2t=0
-2t=-3
2t=3
t=3/2
therfore max displacement is when t=3/2
s=2+3(3/2)-(3/2)^2
s=2+9/2-9/4
s=13/2-9/4
s=17/4
s=4.25m
8. y = x^3 -12x - 12
ReplyDeletedy/dx=3x2-12
3x^2-12=0 ( at st pt =o)
3x^2=12
x^2=4
x=sqrt4
x=2
at x=2 y=2^3 -12(2) - 12
y=8-24-12
-28
therefore the stationary point is x=2 y=-28
green lantern you did number (4)incorrect
ReplyDeletey=x^2+1/x
y=x^2+x^-1
dy/dx=2x-1x^-2
oh yes i did not see the division sign....
ReplyDelete4.
ReplyDeleteFind the derivative of y = x^2 + 1/x
y=2x2-1+1x-1
y=2x+1
ques 9 i
ReplyDelete2x^2 + x + 15
(2x + )( x + )
cannot be factorised
question 4
ReplyDeletey=x^2 + 1/x
y=x^2 + x^-1
dy/dx=2x - x^-2
dy/dx=2x - 1/x^2
question 7
ReplyDeletes=2 + 3t - t^2
ds/dt=3 - 2t
v=3 - 2t
for maximum displacement v=0 therefore
v=3 - 2t
3 - 2t=0
2t=3
t=3/2
question 5
ReplyDeletey=1 + 4x^3
dy/dx=12x^2
question#4
ReplyDeletey=x^2+1/x
y=x^2+x^-1
dy/dx=2x-1x^-2
ques 9
ReplyDeletef(x)= 2x^2 + x + 15
not sure how to factorize
Since:
ReplyDeletey = x^3 -12x - 12
then:
dy/dx=3x^2-12
3x^2-12=0
3x^2=12
x^2=4
x=√4 = 2
when
x=2 y=2^3 -12(2) - 12
y=8-24-12 = -28
so x = 2 and y = -28
since:
ReplyDeletey=1 + 4x^3
dy/dx=(3)4x^3-1
dy/dx=12x^2
7) s=2+3t-t^2
ReplyDeletediff. to find velocity
ds/dt = 3-2t
v= 3-2t, for max disp. vel=0;
max disp:
0= 3-2t
2t= 3
t= 3/2,
at max disp t=3/2
sub t into disp eq'n to find max disp
s=2+3(3/2)-(3/2)^2
s=2+(9/2)-(9/4)
s=4.25
for ques 9(ii)
ReplyDeletef(x)= 2x^2 + x + 15
2( x^2 + x/2 + (1/4)^2 ) + 15 - 2(1/4)^2
2( x + 0.25 )^2 + 14.875
for ques 4
ReplyDeletey=x^2+1/x
y=x^2+x^-1
dy/dx=2x-1x^-2
question 9 cannot be factorised so don't waste your time trying to figure it out .
ReplyDelete(4)y= x^2 + 1/x
ReplyDeletedy/dx= 2x^2-1 + 1x^-1-1
dy/dx= 2x^1 + -x^-2
dy/dx= 2x^1 + -1/x^2
dy/dx= 2x^1 + -1/x^2
(5)
ReplyDeletey= 1 + 4x^3
dy/dx= (3*4)x^(3-1)
dy/dx= (12)x^(2)
dy/dx= 12x^2
# Find the derivative of y = 1 + 4x^3
ReplyDelete# Find the derivative of y = x^2 + 54/x
(6) y= x^2 + 54/x
y= x^2 + 54x^-1
dy/dx= (2*1)x^(2-1) + (54*-1)x^(-1-1)
dy/dx= 2x - 54x^-2
dy/dx= 2x - 54/x^2
(8)
ReplyDeletey= x^3 - 12x - 12 (1)
dy/dx= (1*3)x^(3-1) - (12*1)x^(1-1)
dy/dx= 3x^2 - 12
the stationary point is where the gradient (dy/dx) of the curve is equal to zero
therefore substitute dy/dx=0
0= 3x^2 - 12
12= 3x^2
12/3= x^2
4= x^2
√4= x
x=+/-2
sub x=2 into (1) sub x= -2 into (1)
y= x^3 - 12x - 12 y= x^3 - 12x -12
y= (2)^3 - 12(2)- 12 y= (-2)^3 - 12(-2) -12
y= 8 - 24 -12 y= 8 + 24 -12
y= -28 y= 20
the coordinates of the stationary point are:- when x=2, y=-28 and when x=-2, y= 20
(4)
ReplyDeletey = x^2 + 1/x
DY/DX = 2x^2-1 + 1x^-1-1
DY/DX = 2x^1 + -x^-2
DY/DX = 2x^1 + -1/x^2
DY/DX = 2x^1 + -1/x^2
5)
ReplyDeleteY = 1 + 4x^3
DY/DX = (3*4)x^(3-1)
DY/DX = (12)x^(2)
DY/DX = 12x^2
(2) the vector has an angle of elevation of 50o, the magnitude is 200m.
ReplyDelete=180- (90+ 50)
angle B=40
angle A=50
angle C=90
(7)
ReplyDeletemaximum displacement ds/dt=0
s=2+3t-t^2
ds/dt=3-2t
mkae the equation equal to zero
3-2t=0
-2t=-3
2t=3
t =3/2
maximum displacement is t=3/2
s= 2+3(3/2)-(3/2)^2
s= 2+9/2-9/4
s= 13/2-9/4
s= 17/4
s=4.3
6)
ReplyDeleteY = x^2+54/x
Y = x^2+54x^-1
DY/DX = (2*1)x^(2-1)+(54*-1)x^(-1-1)
DY/DX = 2x-54x^-2
DY/DX = 2x-54/x^2
in order to find the magnitude of the horizontal.. we can use the formula:
ReplyDeletecosᶿ=adj/hyp
cos53= h/200
h= cos53 * 200
h= 120.36
the horizontal = 120.36m
(2)
ReplyDeletein order to find the vertical component of the vector:-
sinᶿ= opp/hyp
sin50= v/ 200
v=sin50*200
v=153.21
therefore the vertical magnitude of the vector= 153.21m
(8)
ReplyDeleteY= x^3 - 12x - 12 (1)
DY/DX = (1*3)x^(3-1) - (12*1)x^(1-1)
DY/DX = 3x^2 - 12
at stationary point the gradient is equal to zero. in other words dy/dx is =0
substitute dy/dx=0
3x^2 - 12 =0
12= 3x^2
12/3= x^2
4= x^2
square root of 4 = x
therfore x= + or - 2
substitute x=2 in to (1) subtitute x= -2 in to (1)
y= x^3 - 12x - 12 y= x^3 - 12x -12
y= (2)^3 - 12(2)- 12 y= (-2)^3 - 12(-2) -12
y= 8 - 24 -12 y= 8 + 24 -12
y= -28 y= 20
stationary point are: when x=2 y=-28 and x=-2 y= 20
can somone work number 9 plz. :)
ReplyDelete8.
ReplyDeletey = x^3 -12x - 12
dy/dx=(1*3)x^2-(12*1)
dy/dx=3x^2 - 12
then dy/dx=0
0=3x^2 - 12
12= 3x^2
12/3= x^2
4= x^2
√4= x
x=+/-2
then sub.x=(2) and x=(-2)
y= x^3 - 12x - 12 y= x^3 - 12x -12
y= (2)^3 - 12(2)- 12 y= (-2)^3 - 12(-2) -12
y= 8 - 24 -12 y= 8 + 24 -12
y= -28 y= 20
1)
ReplyDeletelet
x^2 + 4y^2 = 20....(1)
xy = 4...(2)
from 2
x=4/y..(3)
sub 3 into 1
(4/y)^2+4y^2=20
16/y^2+4y2=20
16/y^2=20-4y^2
16=y^2(20-4y^2)
16=20y^2-4y^4
4y^4-20y^2+16=0
/4
y^4-5Y^2+4=0
Factorize
(y^2+4)(y^2-1)
Y^2=-4 or y^2=1
y=sqrt(-4) or y= sqrt 1
does not exist or +/- 1
therfore sub y=1 into (2)
xy=4
1x=4
x=4
or y=-1
xy=4
-1x=4
x=-4
when y=1 x=4
y=-1 x=-4
7)
ReplyDeletes=2 + 3t - t^2
ds/dt=3 - 2t
v=3 - 2t
for maximum displacement v=0 therefore
v=3 - 2t
3 - 2t=0
2t=3
t=3/2
derivative:
ReplyDeletey = x^2 + 1/x
y=2x2-1+1x-1
y=2x+1
find the derivative:
ReplyDeletey = x^2 + 54/x
y=2x2-1+54x-1
y=2x+54
s = 2 + 3t - t^2
ReplyDeletefor the max displacement ds/dt=0
s=2+3t-t^2
ds/dt=3-2t
hence: 3-2t=0
-2t=-3
2t=3
t=3/2
hence max displacement is t=3/2
s=2+3(3/2)-(3/2)^2
s=2+9/2-9/4
s=13/2-9/4
s=17/4
s=4.25m
y = x^3 -12x - 12
ReplyDeletecord of stationary pt:
dy/dx=3x2-12
3x^2-12=0 ( at st pt =o)
3x^2=12
x^2=4
x=sqrt4
x=2
at x=2 y=2^3 -12(2) - 12
y=8-24-12
-28
x=2 and y=-28
number 1
ReplyDeletex^2 + 4y^2 = 20.....eq1
xy = 4.....eq2
from eq2
x=4/y.....eq3
sub 3 into 1
(4/y)^2+4y^2=20
16/y^2+4y2=20
16/y^2=20-4y^2
16=y^2(20-4y^2)
16=20y^2-4y^4
4y^4-20y^2+16=0
(divide everything by 4because they all are multiples of 4)
y^4-5Y^2+4=0
(y^2+4)(y^2-1)
Y^2=-4 or y^2=1
y=sq.root(-4) or y= sq.root 1
does not exist or +/- 1
therfore sub y=1 into eq2
xy=4
1x=4
x=4
or
y=-1
xy=4
-1x=4
x=-4
when y=1 x=4
y=-1 x=-4
This comment has been removed by the author.
ReplyDelete9 (i)
ReplyDelete2x^2 + x + 15
(2x + )( x + )
its either i can't factorised it or it can't be factorised
9 (ii)
ReplyDeletef(x) = 2x^2 + x + 15
2( x^2 + x/2 + (1/4)^2 ) + 15 - 2(1/4)^2
2( x + 0.25 )^2 + 14.875
number (10)
ReplyDeletey^2 = 12x.... eq1
3y = 4x + 6.....eq2
from eq2
y = (4x + 6)/3.....eq3
substitute eq3 in eq1
( (4x/3) + 2)( (4x/3) + 2) = 12x
(16x^2/9) + (16x/3) + 4 = 12x
multiply by 9 throughout to get rid of the denominators
16x^2 + 48x + 36 = 108x
divide by 4
4x^2 + 12x - 27x + 9 = 0
4x^2 - 15x + 9 = 0
4x^2 - 12x - 3x + 9 = 0
4x(x - 3) -3(x - 3) = 0
(x - 3) (4x - 3) = 0
x = 3 or x = 3/4
sub x = 3 in eqn 3
y = (4 * 3 + 6 )/3
6
sub x = 3/4 in eqn 3
y = ( 4* 3/4 +6 )/3
3
pts of intersection ( 3 , 6) and ( 3/4 , 3)
distance between pts =
( (6 - 3 )^2 - ( 3 - 3/4 )^2 )^0.5
3.75
ques 10
ReplyDeletey^2 = 12x (1)
3y = 4x + 6 (2)
from eqn 2
y = (4x + 6)/3 (3)
substitute eqn 3 in eqn 1
( (4x/3) + 2)( (4x/3) + 2) = 12x
(16x^2/9) + (16x/3) + 4 = 12x
multiply by 9 throughout
16x^2 + 48x + 36 = 108x
/ by 4
4x^2 + 12x - 27x + 9 = 0
4x^2 - 15x + 9 = 0
4x^2 - 12x - 3x + 9 = 0
4x(x - 3) -3(x - 3) = 0
(x - 3) (4x - 3) = 0
x = 3 or x = 3/4
sub x = 3 in eqn 3
y = (4 * 3 + 6 )/3
6
sub x = 3/4 in eqn 3
y = ( 4* 3/4 +6 )/3
3
points of intersection ( 3 , 6) and ( 3/4 , 3)
distance between points =
( (6 - 3 )^2 - ( 3 - 3/4 )^2 )^0.5
3.75
ques 9 i
ReplyDelete2x^2 + x + 15
(2x + )( x + )
cannot be factorised
9 ii
ReplyDeletef(x) = 2x^2 + x + 15
2( x^2 + x/2 + (1/4)^2 ) + 15 - 2(1/4)^2
2( x + 0.25 )^2 + 14.875
9 iii
ReplyDeletef(x) = 2(x + 0.25 )^2 + 14.875
Turning point (-0.25 , 14.875)
9 iv
ReplyDeletey = 2x^2 + x + 15 (1)
dy/dx = 4x + 1
at turning point dy/dx = 0
4x + 1 = 0
x = -0.25
sub x = -0.25 in eqn 1
y = 2*(-0.25)^2 - 0.25 + 15
14.875
turning point ( -0.25,14.875)
9 v
ReplyDeleteYou can just look at the completed square in iii and determine the turning point but in the differential you have to equate dy/dx to zero and then sub the x value in the eqn with y . It is much more longer.
8.
ReplyDeletey = x^3 -12x - 12
cordinates of stationary pt
dy/dx=3x2-12
3x^2-12=0 ( at st pt =o)
3x^2=12
x^2=4
x=sqrt4
x=2
at x=2 y=2^3 -12(2) - 12
y=8-24-12
-28
x=2 y=-28
7.
ReplyDeleteDisplacement is given by s = 2 + 3t - t^2, find the maximum displacement
for max displacement ds/dt=0
so
s=2+3t-t^2
ds/dt=3-2t
therdore 3-2t=0
-2t=-3
2t=3
t=3/2
therfor max displacement is when t=3/2
s=2+3(3/2)-(3/2)^2
s=2+9/2-9/4
s=13/2-9/4
s=17/4
s=4.25m
6.
ReplyDeleteFind the derivative of y = x^2 + 54/x
y=2x2-1+54x-1
y=2x+54
5.
ReplyDeleteFind the derivative of y = 1 + 4x^3
y=1-1+ 4(3)x3-1
y=0+12x2
y=12x^2
4.
ReplyDeleteFind the derivative of y = x^2 + 1/x
y=2x2-1+1x-1
y=2x+1
x^2 + 4y^2 = 20....(1)
ReplyDeletexy = 4...(2)
from 2
x=4/y..(3)
sub 3 into 1
(4/y)^2+4y^2=20
16/y^2+4y2=20
16/y^2=20-4y^2
16=y^2(20-4y^2)
16=20y^2-4y^4
4y^4-20y^2+16=0
/4
y^4-5Y^2+4=0
Factorize
(y^2+4)(y^2-1)
Y^2=-4 or y^2=1
y=sqrt(-4) or y= sqrt 1
does not exist or +/- 1
therfore sub y=1 into (2)
xy=4
1x=4
x=4
or y=-1
xy=4
-1x=4
x=-4
therefore y=1, 4 and x=-1, -4
s=2 + 3t - t^2
ReplyDeleteds/dt=3 - 2t
v=3 - 2t
for maximum displacement v=0 hence
v=3 - 2t
3 - 2t=0
2t=3
t=3/2
y = x^3 -12x - 12
ReplyDeletedy/dx=(1*3)x^2-(12*1)
dy/dx=3x^2 - 12
then dy/dx=0
0=3x^2 - 12
12= 3x^2
12/3= x^2
4= x^2
√4= x
x=+/-2
then sub.x=(2) and x=(-2)
y= x^3 - 12x - 12 y= x^3 - 12x -12
y= (2)^3 - 12(2)- 12 y= (-2)^3 - 12(-2) -12
y= 8 - 24 -12 y= 8 + 24 -12
y= -28 y= 20
y=x^2 + 1/x
ReplyDeletey=x^2 + 1x^-1
dy/dx=2x - 1x^-2
ques 5:
ReplyDeletey=1 + 4x^3
dy/dx= 0 + 12x^2
ques 6:
ReplyDeletey = x^2 + 54/x
y = x^2 + 54x^-1
dy/dx = 2x^1 - 54x^-2
ques 8:
ReplyDeletey = x^3 -12x - 12
dy/dx = 3x^2 - 12x^0 - 0
dy/dx = 3X^2 - 12
ques 8 cont'd:
ReplyDeleteFor stationary point there is no gradient hence dy/dx is 0..
ques 8 cont'd:
ReplyDeleteif dy/dx = 0
and dy/dx = 3x^2 - 12
Then 0 = 3x^2 - 12
ques 8 cont'd:
ReplyDelete0 = 3x^2 - 12
3x^2 = 12
x^2 = 12/3
x = √12/3
x = 2
Therefore coordinate of stationary point :
(2,0)
let
ReplyDeletex^2 + 4y^2 = 20....(1)
xy = 4...(2)
from 2
x=4/y..(3)
sub 3 into 1
(4/y)^2+4y^2=20
16/y^2+4y2=20
16/y^2=20-4y^2
16=y^2(20-4y^2)
16=20y^2-4y^4
4y^4-20y^2+16=0
/4
y^4-5Y^2+4=0
Factorize
(y^2+4)(y^2-1)
Y^2=-4 or y^2=1
y=sqrt(-4) or y= sqrt 1
does not exist or +/- 1
therfore sub y=1 into (2)
xy=4
1x=4
x=4
or y=-1
xy=4
-1x=4
x=-4
when y=1 x=4
y=-1 x=-4
7.
ReplyDeleteDisplacement is given by s = 2 + 3t - t^2, find the maximum displacement
for max displacement ds/dt=0
so
s=2+3t-t^2
ds/dt=3-2t
therdore 3-2t=0
-2t=-3
2t=3
t=3/2
therfor max displacement is when t=3/2
s=2+3(3/2)-(3/2)^2
s=2+9/2-9/4
s=13/2-9/4
s=17/4
s=4.25m
8.
ReplyDeletey = x^3 -12x - 12
cordinates of stationary pt
dy/dx=3x2-12
3x^2-12=0 ( at st pt =o)
3x^2=12
x^2=4
x=sqrt4
x=2
at x=2 y=2^3 -12(2) - 12
y=8-24-12
-28
x=2 y=-28
4.
ReplyDeleteFind the derivative of y = x^2 + 1/x
y=2x2-1+1x-1
y=2x+1
5.
ReplyDeleteFind the derivative of y = 1 + 4x^3
y=1-1+ 4(3)x3-1
y=0+12x2
y=12x^2
6.
ReplyDeleteFind the derivative of y = x^2 + 54/x
y=2x2-1+54x-1
y=2x+54
7.
ReplyDeleteDisplacement is given by s = 2 + 3t - t^2, find the maximum displacement
for max displacement ds/dt=0
so
s=2+3t-t^2
ds/dt=3-2t
therdore 3-2t=0
-2t=-3
2t=3
t=3/2
therfor max displacement is when t=3/2
s=2+3(3/2)-(3/2)^2
s=2+9/2-9/4
s=13/2-9/4
s=17/4
s=4.25m
7.
ReplyDeleteDisplacement is given by s = 2 + 3t - t^2, find the maximum displacement
for max displacement ds/dt=0
so
s=2+3t-t^2
ds/dt=3-2t
therdore 3-2t=0
-2t=-3
2t=3
t=3/2
therfor max displacement is when t=3/2
s=2+3(3/2)-(3/2)^2
s=2+9/2-9/4
s=13/2-9/4
s=17/4
s=4.25m
8.
ReplyDeletey = x^3 -12x - 12
cordinates of stationary pt
dy/dx=3x2-12
3x^2-12=0 ( at st pt =o)
3x^2=12
x^2=4
x=sqrt4
x=2
at x=2 y=2^3 -12(2) - 12
y=8-24-12
-28
x=2 y=-28
number (1)
ReplyDeletelet
x^2 + 4y^2 = 20....(1)
xy = 4...(2)
from 2
x=4/y..(3)
sub 3 into 1
(4/y)^2+4y^2=20
16/y^2+4y2=20
16/y^2=20-4y^2
16=y^2(20-4y^2)
16=20y^2-4y^4
4y^4-20y^2+16=0
/4
y^4-5Y^2+4=0
Factorize
(y^2+4)(y^2-1)
Y^2=-4 or y^2=1
y=sqrt(-4) or y= sqrt 1
does not exist or +/- 1
therfore sub y=1 into (2)
xy=4
1x=4
x=4
or y=-1
xy=4
-1x=4
x=-4
when (4,1),(-4,-1)
number (4)
ReplyDeleteFind the derivative of y = x^2 + 1/x
y=2x2-1+1x-1
y=2x+1
number (5)
ReplyDeleteFind the derivative of y = 1 + 4x^3
y=1-1+ 4(3)x3-1
y=0+12x2
y=12x^2
number (6)
ReplyDeleteFind the derivative of y = x^2 + 54/x
y=2x2-1+54x-1
y=2x+54
number (7)
ReplyDeleteDisplacement is given by s = 2 + 3t - t^2, find the maximum displacement
for max displacement ds/dt=0
so
s=2+3t-t^2
ds/dt=3-2t
therdore 3-2t=0
-2t=-3
2t=3
t=3/2
therfor max displacement is when t=3/2
s=2+3(3/2)-(3/2)^2
s=2+9/2-9/4
s=13/2-9/4
s=17/4
s=4.25m
number (8)
ReplyDeletey = x^3 -12x - 12
co-ord of stat pt
dy/dx=3x2-12
3x^2-12=0 ( at st pt =o)
3x^2=12
x^2=4
x=sqrt4
x=2
at x=2 y=2^3 -12(2) - 12
y=8-24-12
-28
x=2 y=-28
number (10)
ReplyDeletey^2 = 12x (1)
3y = 4x + 6 (2)
from eqn 2
y = (4x + 6)/3 (3)
substitute eqn 3 in eqn 1
( (4x/3) + 2)( (4x/3) + 2) = 12x
(16x^2/9) + (16x/3) + 4 = 12x
multiply by 9 throughout
16x^2 + 48x + 36 = 108x
/ by 4
4x^2 + 12x - 27x + 9 = 0
4x^2 - 15x + 9 = 0
4x^2 - 12x - 3x + 9 = 0
4x(x - 3) -3(x - 3) = 0
(x - 3) (4x - 3) = 0
x = 3 or x = 3/4
sub x = 3 in eqn 3
y = (4 * 3 + 6 )/3
6
sub x = 3/4 in eqn 3
y = ( 4* 3/4 +6 )/3
3
points of intersection ( 3 , 6) and ( 3/4 , 3)
distance between points =
( (6 - 3 )^2 - ( 3 - 3/4 )^2 )^0.5
=3.75
1.
ReplyDeletex^2 + 4y^2 = 20______(1)
xy = 4_______________(2)
2: X = 4/y___________(3)
subst 3 into 1:
(4/y)^2 + 4y^2 = 20
16/y^2 + 4y^2 = 20
16/y^2 = 20 - 4y^2
16 = y^2(20 - 4y^2)
16 = 20y^2 - 4y^4
20y^2 - 4y^4 = 16
20y^2 - 4y^4 - 16 = 0
/4: 5y^2 - y^4 - 4= 0
-y^4 + 5y^2 - 4 = 0
-y^4 + y^2 + 4y^2 - 4 = 0
y^2 - y^4 - 4 + 4y^2 = 0
y^2 (1 - y^2) - 4 (1 - y^2) = 0
(1 - y^2)(y^2 - 4) = 0
1 - y^2 = 0
-y^2 = -1
y^2 = 1
y = +1 or -1
y^2 - 4 = 0
y^2 = 4
y = +2 or -2
subst y values into eqn 2:
when y = +1
xy = 4
x(+1) = 4
x = +4
when y = -1
xy = 4
x(-1) = 4
x = -4
when y = +2
xy = 4
x(+2) = 4
x = +2
when y = -2
xy = 4
x(-2) = 4
x = -2
so corresponding values are as follows:
when(x = +4,y = +1)
when(x = -4,y = -1)
when(x = +2,y = +2)
when(x = -2,y = -2)
4.
ReplyDeletey = x^2 + 1/x
y = x^2 + 1x^-1
dy/dx = 2x - 1x^-2
dy/dx = 2x - 1/x^2
5.
ReplyDeletey = 1 + 4x^3
dy/dx = 12x^2
6.
ReplyDeletey = x^2 + 54/x
y = x^2 + 54x^-1
dy/dx = x - 54x^-2
dy/dx = x - 54/x^2
7.
ReplyDeletes = 2 + 3t - t^2
ds/dt = 3 - 2t
max displacement is when ds/dt = 0
∴ 3 - 2t = 0
-2t = -3
t = 3/2
so max occurs when t = 3/2
subst t value back into original eqn:
s = 2 + 3t - t^2
s = 2 + 3(3/2) - (3/2)^2
s = 4.25m
8.
ReplyDeletey = x^3 -12x - 12
dy/dx = 3x^2 - 12
for stationary point dy/dx = 0
∴ 3x^2 - 12 = 0
3x^2 = 12
x^2 = 4
x = +2 or -2
subst x values into original equation:
when x = +2
y = x^3 -12x - 12
y = (+2)^3 - 12(+2) - 12
y = -28
when x = +2
y = x^3 -12x - 12
y = (-2)^3 - 12(-2) - 12
y = 4
so corresponding values are as follows:
when(x = +2,y = -28)
when(x = -2,y = +4)
question 6
ReplyDeletey=x^2+54/x
y=x^2+54x^-1
dy/dx=2x-54x^-2
=2x-54/x^2
questio 8
ReplyDeletey=x^3-12x-12
dy/dx=3x^2-12
for turning points dy/dx=0
3x^2-12=0
3x^2=12
x^2=4
x=2
substitute x=2 in y=x^3-12x-12
y=(2)^3-12(2)-12
y=8-24-12
y=-28
turning point (2,-28)
question 2
ReplyDeletecos50=adj/200
horizontal component=200cos50
tan50=opp/200
vertical component=200tan50