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

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SUMMARY

The discussion focuses on sketching the graph of the function y = sin(x)(1 - cos(x)) over the interval [-2π, 2π]. Participants identify the first derivative as y' = cos(x)(1 - cos(x)) + sin(2x) and emphasize the necessity of applying the product rule correctly. The critical points for maxima and minima are discussed, with specific values of x identified as π/2, 3π/2, 0, π, and 2π. The correct values for maxima and minima are noted as 2π/3 and 4π/3, respectively, requiring further exploration of the cosine function.

PREREQUISITES
  • Understanding of trigonometric functions, specifically sine and cosine.
  • Familiarity with calculus concepts, including derivatives and the product rule.
  • Ability to solve equations involving trigonometric identities.
  • Knowledge of graphing techniques for periodic functions.
NEXT STEPS
  • Review the product rule in calculus for differentiating products of functions.
  • Learn how to apply trigonometric identities, such as sin(2x) = 2sin(x)cos(x).
  • Study the behavior of the cosine function to identify critical points and their significance.
  • Practice sketching graphs of trigonometric functions over specified intervals.
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and graphing trigonometric functions, as well as educators seeking to clarify concepts related to product differentiation.

Glissando
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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!
 
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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:

(fg)^\prime=f^\prime g+fg^\prime
 
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:

(fg)^\prime=f^\prime g+fg^\prime

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.
 

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