y = 4 has no change since 4 is a constant
y = 4x is a straight line which means steady or fixed change y = mx + c so fixed change is gradient which is m which is 4
y = 4x^2 means this is a curve so there is change at every point(different grad at every point) so calculas helps in approximating by giving the following rate of change
- y=ax^n dy/dx = anx^(n-1)
- y = sin x dy/dx = cos x
- y = cos x dy/dx = - sin x
If gradient or tangent or normal is to be found, it must be at a specific point since each point has its own gradient or tangent or normal.
Polynomial differentiation
- Find dy/dx of the equation y = 2x + 8/(x^2). Find the coordinates of the turning point, is it max or min and verify your answer. Find the normal to the curve AT THE POINT (-2, -2)
- A curve has equation y = k/x where k is a constant. Given that the gradient of the curve is -3 when x = -2, find the value of the constant k.
- The equation of a curve is y = x + 2cos x. Find the x-coordinates of the stationary points of the curve and determine the nature of the turning points.
- The equation of a curve is y = x^3 - 8. Find the equation of the normal to the curve AT THE POINT where the curve crosses the x-axis.
- A curve is such that dy/dx = 16/x^3 and (1, 4) is a point on the curve. Find the equation of the curve.
- A curve is such that dy/dx = 2x^2 - 5 and (3, 8) is a point on the curve. Find the equation of the curve.
Integration deals with area. Area must be bounded otherwise area cannot be found. Hence to be bounded, normally 4 lines must be known to give an enclosed region. Integration of a curve gives the area under the curved trapped towards the x-axis. Two other lines are given by the lines x = 7 to x = 1 which is called upper bound and lower bound. WHen 2 curves are given the bounds are found by determining the points of intersection.
- The curve y = 3x^.5 and the line y = x, find the two x bounds (ie the points of intersection) Find the area under the curve but above the line.
- The region P is bounded by the curve y = 5x - x^2, the x=axis and the line x = h. The region Q is bounded by the curve y = 5x - x^2, the xaxis and the lines x = h and x = 2h. Given that the area of Q is twice the area of P, find the value of h.
1
ReplyDeletey = 2x + 8/x^2
2x + 8(x^-2)
dy/dx = 2 - 16 (x^-3)
1
ReplyDeleteAt turning point dy/dx=0
2 - 16 /x^3 = 0
16/x^3 = 2
16 = 2* x^3
x^3 = 8
x = 2
sub x = 2 in y = 2x + 8/x^2
y = 6
coordinates turning point (2,6)
1
ReplyDeleted2y/dx^2 = 48(x^-4)
It is therefore a minimum point because d2y/dx^2 > 0
1
ReplyDeleteGradient at x = -2
dy/dx = 2 - 16(-2^-3)
4
Gradient of normal at (-2,-2) =
-1/4
eqn of normal
ReplyDeletey - -2 = -1/4(x - -2)
y = (-1/4)x + 3/2
2
ReplyDeletey = k/x
dy/dx = -k*x^-2
2
ReplyDeleteIf the gradient is -3 when x = -2 then
-k*(-2^-2) = -3
-k = -12
k = 12
6
ReplyDeletedy/dx = 2x^2 - 5
y = 2x^3/3 - 5x + C (constant of integration)
6
ReplyDelete(3,8) is a point on the curve
sub (3,8) in y
8 = 3 + C
C = 5
6
ReplyDeletetherefore if C = 5 then the eqn of the curve is
y = (2x^3/3) - 5x + 5
integration 1
ReplyDeletey = 3x^5 intersects the line y = x
therefore
3x^5 = x
3x^4 = 1
x^4 = 1/3
x = 0.76
For question 1 on the integration how do find the points of intersection Please help
ReplyDelete3
ReplyDeletey = x + 2cosx
dy/dx = 1 - 2sinx
Diff question 1)
ReplyDeletey = 2x + 8/(x^2)
y = 2x + 8x^-2
dy/dx = 2 + (-2)(8)x^(-2-1)
dy/dx = 2 - 16x -3
at point (-2,2);
dy/dx = 2 - 16(-2)^-3
dy/dx = 2 - 16(-8)
dy/dx = 2 - (-128)
dy/dx = 2 + 128
dy/dx = 130
finding the normal to the curve:
the normal is a straight line and thus has the equation y = mx + c
m is the gradient and since dy/dx is also gradient, m = dy/dx = 130.
from the coordinates given, y = 2.
therefore, 2 = 130(-2) + c
2 = - 260 + c
c = 2 + 260
c = 280
therefore the equation of the normal is
y = 130x + 280
This comment has been removed by the author.
ReplyDelete#1.y = 2x + 8/(x^2)
ReplyDeletedy/dx= 2 - 16/(x^3)
At turning pt. dy/dx=0
0= 2 - 16/(x^3)
x= 2, y= 6
d2y/dx2= 48/(x^4)
subs x=2
d2y/dx2=3
therefore it is a minimum point
Normal at point (-2,2)
dy/dx= 4
Therfore at normal gradient = -1/4
y= mx + c
2= (-1/4)(-2) + c
2= 1/2 + c
c = 3/2
equation of normal = 4y = -x + 6
#2. y = k/x
ReplyDeletedy/dx = -k/(x^2)
where x = -2, dy/dx = -3
therfore
-3 = -k/4
k = 12
y = x + 2cosx
ReplyDeletedy/dx = 1 - 2sinx
At staionary point dy/dx = 0
i.e. 1 = 2sinx
x = pi/6 rad
d2y/dx2 = -2cosx
subs x = pi/6
d2y/dx2 = negative
therefore point is a maximum
#4
ReplyDeletey = x^3 - 8
dy/dx = 3x^2
#4
ReplyDeleteWhere curve crosses x axis, y = 0
x^3 = 8
x = 2
#4
ReplyDeleteWhere x = 2
d2y/dx2 = 6x
= 12
Therefore point is a minimum
dy/dx = 16/x^3 and point (1,4)
ReplyDeleteintegrating
y = -8/x^2 + c
#4
ReplyDeleteAt point (1,4)
4 = -8/1 + c
c = 12
y = -8/x^2 + 12
dy/dx = 2(x^2) - 5
ReplyDeleteintegrating
y = 2(x^2)/3 - 5x + c
#6 contd
ReplyDeleteAt point (3,8)
8 = 18 - 15 + c
c = 5
y = 2(x^2)/3 - 5x + 5
y=2x+8/(x^2)
ReplyDeletey=2x+8x^-2
dy/dx=2+(-2)(8)x^(-2-1)=2-16x^-3
at turning point dy/dx=0
2-16/(x^3)=0
16/(x^3)=2
16=2*x^3
x^3=8
x=2
substitute x=2 into y=2x+8/(x^2)=4+2=6
(x,y) = ((2,6)
d2y/dx^2=48/x^4
subst. x=2
dy/dx=3 minimum point
normal at (-2,-2)
dy/dx=2-16(-2^-3)
dy/dx=4
gradient of normal =-1/4
#2
ReplyDeletey=k/x
dy/dx=-k*x^-2
when x=-2 an gradient=-3,
-k*(-2^-2)=-3
k=12
3 ii
ReplyDeleteat a turning point dy/dx = 0
1 - 2sinx = 0
sinx = 1/2
x = 30 , 150 (degrees)
0.52,2.6 ( radians)
3 iii
ReplyDeletedy/dx = 1 - 2sinx
d2y/dx^2 = -2cosx
Both max and min points are present
QUESTION OPEN TO ANYONE
ReplyDeleteA curve passes through the point (-2,4) and the gradient of the curve at any point can be calculated using x^2 + 3x + 7 . Find the eqn of the curve.
FOLLOW UP QUESTION
ReplyDeleteFIND THE EQN OF THE TANGENT TO THE CURVE AT THE POINT (-2,4)
THIS IS FREE ADVICE TO ANYONE,WHEN FINDING AREA USING INTEGRATION
ReplyDeleteALWAYS MAKE A SKETCH OF THE CURVE AND THE AREA YOU WANT TO FIND
PS. DON'T FORGET TO PUT IN YOUR LIMITS
CAN SOMEONE PLEASE TELL ME HOW IT IS POSSIBLE TO INTEGRATE BETWEEN TWO X BOUNDARIES AND END UP WITH AN AREA = 0 ????????????????????????
ReplyDelete1) dy/dx=2x+8/x^2
ReplyDelete2x+8x^-2
gradiant of curve=dy/dx
dy/dx= 2-16x^-3
tp=dydx=0
2-16/x^3=0
16-2x^3=
2x^3=16
x=2
min turning point
at the normal (-2,-2)
y=mx+c
-2=-2m+c
m=2-16/-8=4
m=4
normal=-1/4
-2=-1/4(-2)+c
C= -5/2
y=-1/4x-5/2
miss please check over this question
RE:CAN SOMEONE PLEASE TELL ME HOW IT IS POSSIBLE TO INTEGRATE BETWEEN TWO X BOUNDARIES AND END UP WITH AN AREA = 0 ????????????????????????
ReplyDeletei do not think this is quite possible Reaper...you will be using the same equation but substituting the two different values of the limits. unless the two limits are the same, i dont see how it is possible to get an area of zero...anyone else has another view of this?
y = 2x + 8/x^2
ReplyDelete2x + 8(x^-2)
dy/dx = 2 - 16 (x^-3)
At turning point dy/dx=0
ReplyDelete2 - 16 /x^3 = 0
16/x^3 = 2
16 = 2* x^3
x^3 = 8
x = 2
sub x = 2 in y = 2x + 8/x^2
y = 6
coordinates turning point (2,6)
question 1
ReplyDeletey=2x+8/(x^2)
y=2x+8x^-2
dy/dx=2+(-2)(8)x^(-2-1)=2-16x^-3
turning point dy/dx=0
2-16/(x^3)=0
16/(x^3)=2
16=2*x^3
x^3=8
x=2
sub x=2 into y=2x+8/(x^2)=4+2=6
(x,y) = ((2,6)
d2y/dx^2=48/x^4
sub x=2
dy/dx=3 this is a minimum point
normal at (-2,-2)
dy/dx=2-16(-2^-3)
dy/dx=4
gradient of normal =-1/4
question 2
ReplyDeletey=k/x
dy/dx=-k*x^-2
when x=-2 an gradient=-3,
-k*(-2^-2)=-3
k=12
need help with integration questions!!!!!!!
ReplyDeleteThis comment has been removed by the author.
ReplyDeletey=2x+8/(x^2)
ReplyDeletey=2x+8x^-2
dy/dx=2+(-2)(8)x^(-2-1)=2-16x^-3
at turning point dy/dx=0
2-16/(x^3)=0
16/(x^3)=2
16=2*x^3
x^3=8
x=2
substitute x=2 into y=2x+8/(x^2)=4+2=6
(x,y) = ((2,6)
d2y/dx^2=48/x^4
subst. x=2
dy/dx=3 minimum point
normal at (-2,-2)
dy/dx=2-16(-2^-3)
dy/dx=4
gradient of normal =-1/4
y = 2x + 8/(x^2)
ReplyDeletedy/dx= 2 - 16/(x^3)
At turning pt. dy/dx=0
0= 2 - 16/(x^3)
x= 2, y= 6
d2y/dx2= 48/(x^4)
subs x=2
d2y/dx2=3
therefore it is a minimum point
Normal at point (-2,2)
dy/dx= 4
Therfore at normal gradient = -1/4
y= mx + c
2= (-1/4)(-2) + c
2= 1/2 + c
c = 3/2
equation of normal = 4y = -x + 6
2
ReplyDeleteIf the gradient is -3 when x = -2 then
-k*(-2^-2) = -3
-k = -12
k = 12
1) y=2x+8/(x^2)
ReplyDeletey=2x+8x^-2
dy/dx = 2 -16x^-3
Turning point
dy/dx=0
2-16x^-3=0
-16x^-3=-2
x^-3= -2/-16
1/x^3= 1/8
x^3=8
x=2
subs. x into y=2x+8/x^2
y= 2(2)+ 8/2^2
y=4+2
y=6
minimum turning point
d2y/dx^2 = 48x^-4
subs x=2
d2y/dx^2 = 3
at normal(-2,-2)
y=mx+c
-2=-2m+6
-2-6=-2m
m=4
therefore gradient of normal = -1/4
2) y=k/x
ReplyDeletedy/dx=-kx^-2
when gradient = -3,x=-2
-3=-k(-2^-2)
-3=-k(1/4)
-k=-12
k=12
5)
ReplyDeletedy/dx=16/x^3
dy/dx=16x^-3
integration
y=16x^-2/-2+C
at point (1,4)
4=16(1^-2)/-2+C
4=16/-2+c
cross multiply
-8+4C=16
4C=24
C=6
Eq'n
y=16x^-2/-2+6
y=16x^-2/4
6)
ReplyDeletedy/dx=2x^2-5
integration
y=2x^3/3-5x+C
at point (3,8)
8=2(3^3)/3-5(3)+c
8=2(27)/3-15+c
8=18-15+c
c=5
y = 2x+8/x^2
ReplyDelete2x+8(x^-2)
dy/dx = 2-16 (x^-3)
at turn pt.
dy/dx= 0
2 - 16 /x^3 = 0
16/x^3 = 2
16 = 2* x^3
x^3 = 8
x = 2
sub x = 2 into the equation y = 2x+8/x^2
y= 2(2)+ 8/(2)^2
y = 6
nornal of the curve
ReplyDeletedy/dx= 2+16/x^-3
at x=2
dy/dx=4
nrmal gradient= -1/4
y=mx+c
-2=-1/4(-2)+c
-2=1/2+c
-2-1/2=c
-3/2=c
eq of normal
y=-1/4x-3/2
#1.
ReplyDeletey = 2x + 8/(x^2)
dy/dx= 2 - 16/(x^3)
At turning pt. dy/dx=0
0= 2 - 16/(x^3)
x= 2, y= 6
d2y/dx2= 48/(x^4)
subs x=2
d2y/dx2=3
therefore a minimum point
Normal at point (-2,2)
ReplyDeletedy/dx= 4
Therfore at normal gradient = -1/4
y= mx + c
2= (-1/4)(-2) + c
2= 1/2 + c
c = 3/2
equation of normal = 4y = -x + 6
#2
ReplyDeletedy/dx = -k/(x^2)
where x = -2,
dy/dx = -3
-3 = -k/4
k = 12
3
ReplyDeletey = x + 2cosx
dy/dx = 1 - 2sinx
2. y = k/x
ReplyDeletedy/dx = -k/(x^2)
where x = -2, dy/dx = -3
therfore
-3 = -k/4
k = 12
1). y=2x+8/x^2
ReplyDeletey=2x+8x^-2
dy/dx=2-16x^-3
dy/dx=2-16/x^3
for turning pts. dy/dx=0
2-16/x^3=0
-16/x^3=-2
-2x^3=-16
x^3=-16/-2
x^3=8
x=cubrt8
x=2
when x=2
y=2(2)+8/(2)^2
y=4+2
y=6
the turning pts. is 92,6)
d2y/dx2=48x^-1
it has a mim. pt. because d2y/dx2>0
1)(ii). dy/dx=gradient
ReplyDeletedy/dx=2-16/x^3
dy/dx=2-16/(-2)^3
dy/dx=2-(-2)
dy/dx=2+2
dy/dx=4
gradient of the normal is 1/4
y=1/4x+c
now using the pt. (-2,2)
2=1/4(-2)+c
2=-1/2+c
c=5/2
so therefore the equation of the normal is
y=1/4x+5/2
2). y=k/x
ReplyDeletey=kx^-1
dy/dx= -kx^-2
now using dy/dx= -3 and when x= -2
-3= -k/(-2)^2
-3= -k/4
-k= -12
k= 12
miss can u do #1 y=3x^5 and #2 y=5x-x^2 please
ReplyDeletequestion 1
ReplyDeletey=2x+8/x^2
y=2x+8x^-2
dy/dx=2-16x^-3
for turning point dy/dx=0
2-16/x^3=0
2x^3-16=0
2x^3=16
x^3=16/2
x^3=8
x=2
substitute x=2 in y=2x+8/x^2
y=2(2)+8/(2)^2
y=4+8/4
y=4+2
y=6
turning point (2,6)
d^2y/dx^2=48x^-4
when x=2
d^2y/dx^2=48/(2)^4
=48/16
=3 positive d^2y/dx^2 means the curve
is a minimum
gradent of the curve at x=-2
dy/dx=2-16/x^3
=2-16/(-2)^3
=2-(16/-8)
=2-(-2)
=4
gradent of the normal=-1/m
=-1/4
equation of the normal:y-y'=m(x-x')
at the point (-2,2) :y-2=-1/4(x-(-2))
:y-2=-x/4-1/2
:y=-x/4-1/2+2
:y=-x/4+3/2
:4y=-x+6
question 2
ReplyDeletey=k/x
y=kx^-1
dy/dx=-kx^-2
=-k/x^2
when dy/dx=-3, x=-2
-3=-k/(-2)^2
-3=-k/4
-12=-k
k=12
#1
ReplyDeletey=2x+8/x^2
y=2X+8x^-2
dy/dx=2-16x^-3
for ststionary point dy/dx=0
2-16x^-3=0
-16x^-3=-2
x^-3=-2/-16
1/x^3=1/8
x=2
#1
ReplyDeletesub into equ y=2x+8/x^2 when x=2
y=2x+8/(2)^2
y=4+2
y=6
#1
ReplyDeletei.e d^2y/dx^2=2-16x^-3
d^2y/dx^2=48x^-4
d^2y/dx^2=48/x^4
d^2y/dx^2=3
#2
ReplyDeletey=k/x
y=kx^-1
dy/dx=-kx^-2
=-k/x^2
sub when x=-2 & dy/dx=-3
dy/dx=-k/x^2
-3=-k/(-2)^2
-3=-k/4
-k=-3(4)
-k=-12
k=12
#3
ReplyDeletey=x+2cosx
dy/dx=1-2sinx
#3
ReplyDeletewhen dy/dx=0
1-2sinx=0
-2sinx=-1
sinx=-1+2
x=-1+2sin
#4
ReplyDeletey=x^3-8
dy/dx=3x^2
#4
ReplyDeletesub when y=0
o=x^3-8
x^3-8=0
x^3=8
x=2
#4
ReplyDeleted^2y/dx^2=6x
sub when x=2
d^2y/dx^2=6(2)
=12
#5
ReplyDeletedy/dx=16/x^3
dy/dx=16x^-3
integrate
y=16x^-2/-2+c
#6
ReplyDeletedy/dx=2x^2-5
integrate
y=2x^3/3-5x+c
#6
ReplyDeletesub when x=3 & y=8
8=2(3)^3/3-5(3)+c
18-15+c=8
c=8-18+15
c=5
no5. dy/dx = 16/x^3 and (1, 4) fnd curve.
ReplyDelete∫16/x^3
=16x^-2/-2
=-8x^-2 +c
therefore y= -8x^-2 +c
sub (1,4)
4= -8(1)^-2 +c
4= -8 +c
c= 12
there eq is y = -8x^-2 +12
At turning point dy/dx=0
ReplyDelete2 - 16 /x^3 = 0
16/x^3 = 2
16 = 2* x^3
x^3 = 8
x = 2
sub x = 2 in y = 2x + 8/x^2
y = 6
coordinates turning point (2,6)
Question 1 solution
2
ReplyDeletey=k/x
y=kx^-1
dy/dx=-kx^-2
=-k/x^2
sub when x=-2 & dy/dx=-3
Question 2 solution
dy/dx=-k/x^2
-3=-k/(-2)^2
-3=-k/4
-k=-3(4)
-k=-12
k=12
sub when x=3 & y=8
ReplyDelete8=2(3)^3/3-5(3)+c
18-15+c=8
c=8-18+15
c=5
Question 6