Question

A curve has the eqn y = 2sin^2x - 5cos x. Find in terms of p the new value of y as x changes from π/2 to (π/2 - p), where p is small.

Answer 1

\[y=2\sin^2x-5 \cos x\]

Answer 2

hi kira, are you in the rate of change in calculus?

Answer 3

What about it?

Answer 4

is the prob about the rate of change?

Answer 5

I'm not sure... But to me it seems like it's related to limits...

Answer 6

so your class is now in limits or rate of change

Answer 7

This question is posed to me by my friend... I don't know his course though...

Answer 8

ok this is just a derivative
f(x) - f(c) f(pi/2) - f(pi/2 -p)
f '(c)= lim ) -------- = lim -----------------
x->c x - c pi/2->(pi/2 -p) pi/2 - (pi/2 - p)

=2 sin2(pi/2)-5cos(pi/2) - 2 sin2(pi/2 -p)-5cos(pi/2 -p)
---------------------------------------------
pi/2 - (pi/2 - p)
you need to use trig and some identities to arrive in the derivative of f(x) here or

dy/dx = Dx [ 2(sin 2x)^2 ]- 5 cos x]
= 2[4sin2x cos2x] + 5 sinx

Answer 9

Thanks

Answer 10

or maybe the prob may be looking only on increment or change only....
use the formula....
change in y = f(x) - f(c) or use
change in y = f(x+h) - f(x)

,,if h=x-c and x=pi/2 , c=pi/2 -p
h=pi/2 -(pi/2 -p)= p where h=p=small
change in y = f(x+h) - f(x)
change in y = f(x+p) - f(x)
change in y = 2sin^2(x+p) - 5cos (x+p) - [ 2sin^2x - 5cos x]

Answer 11

change in y = 2sin^2(pi/2+p) - 5cos (pi/2+p) - [ 2sin^2x - 5cos x] you need to use trig and some identities to arrive in the change in y or delta y