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

You should find four solutions for x.)In summary, the student was attempting to sketch the graph of y = sinx(1-cosx) over [-2pi, 2pi] and had correctly found the first derivative to be y' = cosx - cos2x + sin2x. However, they made a mistake in simplifying the equation and were unsure of how to solve for the correct values of cosx. Another user pointed out the error and suggested using the product rule to find the first derivative, which led to the correct solution. The student then made a mistake in finding the second derivative, but was able to correct it with the help of another user. In the end, the student was able to
  • #1
Glissando
34
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
  • #2
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]
 
  • #3
Once you get the first derivative right, don't forget to find the second derivative.
 
  • #4
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!
 
  • #5
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.
 

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

1. What is the general shape of the graph of y = sinx(1-cosx)?

The graph of y = sinx(1-cosx) is a sinusoidal curve that oscillates between positive and negative values. It has a period of 2π and an amplitude of 1.

2. How do you determine the x-intercepts of the graph of y = sinx(1-cosx)?

The x-intercepts of the graph occur when sinx(1-cosx) = 0. This happens when either sinx = 0 or 1-cosx = 0. Solving for x in these equations, we get x = 0, π, 2π, and so on.

3. How do you find the maximum and minimum values of the graph of y = sinx(1-cosx)?

The maximum value occurs when sinx = 1 and 1-cosx = 1, which happens at x = π/2 and all multiples of 2π. The minimum value occurs when sinx = -1 and 1-cosx = 1, which happens at x = 3π/2 and all odd multiples of π.

4. Can you sketch the graph of y = sinx(1-cosx) without using a calculator?

Yes, you can sketch the graph using the properties of the sine and cosine functions. First, plot the x-intercepts at x = 0, π, 2π, etc. Then, plot the maximum and minimum points at x = π/2, 3π/2, 5π/2, etc. Finally, connect these points with a smooth curve to complete the graph.

5. What is the relationship between the graphs of y = sinx and y = sinx(1-cosx)?

The graph of y = sinx(1-cosx) is similar to the graph of y = sinx, but it is stretched vertically by a factor of 1-cosx. This means that the amplitude of y = sinx(1-cosx) is greater than the amplitude of y = sinx. Additionally, the x-intercepts and maximum and minimum points of y = sinx are also present in the graph of y = sinx(1-cosx).

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
489
  • Calculus and Beyond Homework Help
Replies
2
Views
764
  • Precalculus Mathematics Homework Help
Replies
11
Views
258
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
393
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
961
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Back
Top