- Find the trap area from (1,4) to (2,1) and the line y = 7 - 3x and the curve y = 4/x^2
- Find the equation of the curve which passes through the point (3,6) and for which dy/dx = 2x(x -3). Hint open brackets first then integrate.
- Differentiate e^(3-2x)
- A particle moves in a straight line so that, t seconds after passing through a fixed point O, its velocity, is giveb by v = 5t -3t^2 + 2. The particle comes to instantaneous rest at the point Q. Find
a) the acceleration of the particle atQ
b) the total distance travelled in the time interval t = 0 to t =3.
y=e^(3-2x)
ReplyDeletelet m= (3-2x)
dm/dx= -2x
rewrite y=e^m
dy/dx= e^m
dy/dx= dy/dm * dm/dx
= e^ (3-2x) * -2x
question3
ReplyDeletee^3-2x
dy/dx=-2e^3-2x
question2
ReplyDeletedy/dx=2x(x-3)
dy/dx=2x^2-6x
y=2/3x^3-3x^2
dy/dx=2x(x-3)
ReplyDeletedy/dx=2x^2-6x
dy/dx=2x^2-6x^2/3=> inttergrate
there for: y=2/3x^3-3x^2
Dark Angel and the best one, when u inttergrate in this case u suppose to end up with a constant C..and u suppose to find this using the points given..well that's what i think.
ReplyDeleteQues 2
ReplyDeletedy/dx=2x(x-3)
expand the brackets
so dy/dx=2x^2-6x
now u integrate.
y=2/3x^3-3x^2 +c
Subst. the points (3,6) to find c
6=18 - 27 + c
c=15
therefore equation of the curve is
y=2/3x^3-3x^2 + 15
Ques 4
ReplyDeleteAt instantaneous rest velocity is zero.
so 5t-+2=0
-3t^2+5t+2=0
(3t + 1)(-t + 2)=0
t=2
a)to find the acc. u have to Differentiate.
so dv/dt= 5 - 6t
subst. t=2
acc= -7ms^-2
b)TO find the total distance travelled u need to integrate the (V) velocity.
v = 5t -3t^2 + 2
y= 5/2t^2 - t^3 + 2t
to find distance, subst.the time interval t = 0 to t =3.
so Dist= 1.5
turbo you are correct dark angel and the best one forgot integration always give + c for constant
ReplyDeleteThat is why the point (3,6) was given so that c can be found
y=2/3x^3-3x^2 + c
x = 3 and y = 6 so c an be found
so when u're integrating, the answer will be:
ReplyDeletey=ax+c
dy/dx = 2x(x -3)
ReplyDeleteWHEN DOING THIS QUESTION OPERN BRACKETS FIRST THEN INTIGRATE
2X^2 - 6X
INTERGRAL 2X^2 - 6X dx
2X^3/3 - 6X^2/2 + c
Can someone plz help me wit #4...
ReplyDelete#4. I think i figured it out.
ReplyDeletev = 5t -3t^2 + 2
a)the acceleration of the particle atQ:
Acceleration is change in velocity/change in time.
Therefore:
acceleration = dv/dt = 5-6t
I do not know how to find the acceleration at Q.
can someone plz help me?
b) the total distance travelled in the time interval t = 0 to t =3.
To find the distance you have to integrate the velocity, because, velocity is the change in distance with respect to time. when you integrate ds/dt you will get the distance moved.
integral of: 5t -3t^2 + 2 dv/dt
s => 5t^2/2 - 3t^3/3 + 2t
then substitute t=3 and t=0 into the s(dist) and minus the lower limit (t=o) from the higher limit (t=3) and you will get your distance moved in the time interval t=o to t=3.
the dist moved is = 1.5
" DARk ANGEL" i think that at point Q the particle should not have any movement becaseu it can at rest at that point ...i have no idea somebody else help???
ReplyDelete