Question

Qn on Geometry. see attached. thx u.

Answer 1

do you have y' ?

Answer 2

actually do you have dy/dx at x = pi/2 ?

Answer 3

btw, this is a calculus question mostly, not geometry.

Answer 4

I don't want to do the whole question either.
Gradient gives you the gradient of the normal to get Q then integrate over 0 to Q on the y

Answer 5

that means get dy/dx = 2/pi + cos x which is the gradient, then use this gradient to get the gradient of the normal using m1 x m2=-1? But i got stuck after getting the gradient of the normal.

Answer 6

If you have "m" and a point on a line, then u know the equation of the line, right?
So then you have Q.

Answer 7

i tried solving but was unable to get the ans Q(0,2+pi^2/4). I tried twice and I still get Q(0, 2+pi^2/(4+2pi)

Answer 8

what's your equation that's normal to the line tangent to the curve at (pi/2, 2) ?

Answer 9

the slope of the line perpendicular to the tangent is -pi/2 . did you use that?

Answer 10

I didn't get the equation instead i use (y-2)/(0-pi/2) = the gradient of normal which is -pi/(2+pi cos 0), had subst all x with 0.

Answer 11

your equation for that line should be
y - 2 = (-pi/2)*(x - pi/2)
set x=0 then solve for y...

Answer 12

why is it the slope of the line perpendicular to the tangent is -pi/2?

Answer 13

because that's the definition of a "normal" to a tangent...