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

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Homework Help Overview

The discussion revolves around sketching the graph of the function y = sinx(1-cosx) over the interval [-2pi, 2pi]. Participants are exploring the necessary derivatives to analyze the function's behavior.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to find the first and second derivatives of the function, discussing the critical points where the derivative equals zero. There is confusion regarding the correct application of the product rule and the resulting expressions for the derivative.

Discussion Status

Some participants have provided guidance on the correct differentiation process, while others are questioning the accuracy of their derivative calculations. Multiple interpretations of the derivative's behavior are being explored, and there is a recognition of the need to clarify the approach to finding critical points.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance they can provide to one another. There is an ongoing discussion about the implications of the derivative results on the graph's features.

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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