[bigbadb1319]

8). A curve has aparametric equatiosn x=t² - 2, y=2t, where -2 greater than and equal to t and less than and equal to 2.
a). Draw a graph of the curve
b). Indicate on your grahp where i. t=0 ii. t>0 iii. t<0
c). Calculate the area of the finite region enclosed by the curve and the y-axis

i've done a and b, can't seem to figure out part c. i've done upto intergrating ydx/dt, but dont' know what the limits are?

thanx...

[Mr. Pink]

enclosed by y axis means, one limit is x = 0, and the other is y = 0, which is t = 0, which means x = -2...

[yusufu]

y = 2t
along x axis, y= 0
2t = 0
t = 0

x = t² -2
x = 0-2
x = -2

limts are 0 and -2

[bigbadb1319]

thanx guys

[bigbadb1319]

umm i jus chked da answer at da back of the book n its 16√2 / 3, n i kinda played around wid da limits n got da √2/3 part but don't know where the 16 comes from?? anyone know wat da correct answer is???

[charikaar]

hi

we have not yet started c4 but i do have the solution disk. let me know the excercise number and i will copy the complet solution here(edexcel ony).

thanks

[bigbadb1319]

Originally Posted by charikaar

hi

we have not yet started c4 but i do have the solution disk. let me know the excercise number and i will copy the complet solution here(edexcel ony).

thanks

thanx a lot, k its C4- Chapter 2: Exercise 2E (Mixed Exercise) Question 8c

[charikaar]

for c hint is:

(c) By symmetry the area above the x-axis is equal to the area below the x-axis

let me type the solution, mouse won't let me copy it....plz bear with me.

--------------

Whenx=−2,t2−2=−2,sot=0
Whenx=0,t2−2=0,sot=√2

dx
dt
=2t
So ydx
dt
=2t×2t=4t2
Therefore A=2∫0√24t2dt
=2é
ë
4
3
t3ù
û
0√2
=2é
ë
4
3
(√2)3−4
3
(0)3ù
û

=2×4
3
(√2)3
=8
3
(√2)3
=16
3
√2, As (√2)3=(√2×√2)×√2=2√2


Does that make sense?

thanks

[bigbadb1319]

Originally Posted by charikaar

for c hint is:

(c) By symmetry the area above the x-axis is equal to the area below the x-axis

let me type the solution, mouse won't let me copy it....plz bear with me.

--------------

Whenx=−2,t2−2=−2,sot=0
Whenx=0,t2−2=0,sot=√2

dx
dt
=2t
So ydx
dt
=2t×2t=4t2
Therefore A=2∫0√24t2dt
=2é
ë
4
3
t3ù
û
0√2
=2é
ë
4
3
(√2)3−4
3
(0)3ù
û

=2×4
3
(√2)3
=8
3
(√2)3
=16
3
√2, As (√2)3=(√2×√2)×√2=2√2


Does that make sense?

thanks

umm i kinda get it....but can u just tell me what the limits they used?? cuz i've done all the begining part jus can't seem 2 figure out da limits u've used...thanx..