Qn. (1 Viewer)

[chousta]

ey all, cant seem to get the 2nd question.


been doin alot of algebra bashing, nothin that gets rid of the 't' term.

tried to solve it using max/min, then also parametrics.


nothing.

http://imageshack.us

 

[jyu]

(i) Find the area of parabolic segment by integration = integ(-1, 2) of x + 2 - x^2 = 9/2

(ii) Area of triangle
= area of trapezium under AB - area of trapezium under AP - area of trapezium under PB
= (1/2)(1 + 4)3 - (1/2)(1 + t^2)(t+1) - (1/2)(t^2 + 4)(2 - t)
=(3/2)(2 + t - t^2)

dA/dt = (3/2)(1 - 2t) =0, t =1/2

Max area = (3/2)(9/4) = (3/4)(9/2)

 

[chousta]

(i) Find the area of parabolic segment by integration = integ(-1, 2) of x + 2 - x^2 = 9/2

(ii) Area of triangle
= area of trapezium under AB - area of trapezium under AP - area of trapezium under PB
= (1/2)(1 + 4)3 - (1/2)(1 + t^2)(t+1) - (1/2)(t^2 + 4)(2 - t)
=(3/2)(2 + t - t^2)

dA/dt = (3/2)(1 - 2t) =0, t =1/2

Max area = (3/2)(9/4) = (3/4)(9/2)

cheers jyu. :wave: