Sketch the graph of y = sinx(1-cosx)

  • Thread starter Thread starter Glissando
  • Start date Start date
  • Tags Tags
    Graph Sketch
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 8K views
Glissando
Messages
34
Reaction score
0

Homework Statement


Sketch the graph of y = sinx(1-cosx) over [-2pi, 2pi]


Homework Equations


first derivative and second derivative


The Attempt at a Solution



y' = cosxsinx
0 = cosxsinx
x = pi/2, 3pi/2, 0, pi, 2pi

I know I have the sin values right (the answers say the same) but the cos values in the answers are actually 2pi/3 for MAX and 4pi/3 for MIN. For that cosx would have to equal -1/2. I'm not sure how to get that );

Thank you!
 
Physics news on Phys.org
Hi Glissando! :smile:

Glissando said:

Homework Statement


Sketch the graph of y = sinx(1-cosx) over [-2pi, 2pi]


Homework Equations


first derivative and second derivative


The Attempt at a Solution



y' = cosxsinx
0 = cosxsinx
x = pi/2, 3pi/2, 0, pi, 2pi

I know I have the sin values right (the answers say the same) but the cos values in the answers are actually 2pi/3 for MAX and 4pi/3 for MIN. For that cosx would have to equal -1/2. I'm not sure how to get that );

Thank you!

Your derivative is wrong. If you want to differentiate a product, then you can't just differentiate both factors and be done with it. You need to apply the product rule:

[tex](fg)^\prime=f^\prime g+fg^\prime[/tex]
 
Once you get the first derivative right, don't forget to find the second derivative.
 
micromass said:
Hi Glissando! :smile:



Your derivative is wrong. If you want to differentiate a product, then you can't just differentiate both factors and be done with it. You need to apply the product rule:

[tex](fg)^\prime=f^\prime g+fg^\prime[/tex]

Good point (wow that was embarassing...)

K well I tried doing it and I still get cosx = 0:

y' = cosx(1-cosx) + sin2x
y' = cosx - cos2x + sin2x
0 = -cosx + cos2x - sin2x
0 = -cosx + cos2x
0 = cosx, x = pi/2, 3pi/2

UGH ): Thanks guys for putting up with me!
 
Well, you have y' right, but ...

-cos(x) + cos(2x) ≠ cos(x)

(Plug x = π/2 into your equation for y' to check your answer.)

You could use sin2(x) =1 - cos2(x), then y' = 0 gives a quadratic in cos(x) . → Factor, or use the quadratic formula to solve for cos(x), and then solve for x.