Have to find the stationary points on the graph y = 3sin^2x

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In summary, the conversation was about finding the solutions for the equation y = 3sin^2x. The conversation went through the steps of finding the derivative and setting it equal to zero to find the stationary points. The conversation also touched on the idea of finding the solutions for sin(2x) = 0 and determining the values of x within the given region. It was also mentioned that there may be other solutions that were not accounted for in the initial answer given by the textbook. Overall, the conversation was focused on finding solutions for the given equation and exploring different methods to do so.
  • #1
Zander Forrester
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Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
y = 3sin^2x
dy/dx = 3(2sinxcosx)
dy/dx = 6sinxcosx
For stationary points dy/dx = 0
6sinxcosx = 0
Can you help me from this point please?
 
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  • #2
You have three factors, 6, sin(x) and cos(x). What must be true for their product to be equal to zero?
 
  • #3
Hello Orodruin,
Thank you for your help.
sin(x) = 0 x = Pi
cos(x) = 0 x = pi/2 and 3pi/2
 
  • #4
Zander Forrester said:
sin(x) = 0 x = Pi
This is one of many solutions.

Zander Forrester said:
cos(x) = 0 x = pi/2 and 3pi/2
These are two of many solutions.

Unless you specify the allowed region of x, you need to account for all solutions.
 
  • #5
Hello again Orodruin,
Region was from 0 to 2pi.
Thank you again for your help.
 
  • #6
Zander Forrester said:

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
y = 3sin^2x
dy/dx = 3(2sinxcosx)
dy/dx = 6sinxcosx
For stationary points dy/dx = 0
6sinxcosx = 0
Can you help me from this point please?
Your 2nd equation can be written as dy/dx = 3(sin(2x)), so it's just a matter of finding where sin(2x) = 0 on the interval [0, 2π].
 
  • #7
Zander Forrester said:
Hello again Orodruin,
Region was from 0 to 2pi.
Thank you again for your help.
So, there are a total of 5 solutions; you have given only three.
 
  • #8
Thank you again.
 
  • #9
Have you ever seen sinxcosx in a formula before?

Maybe you could use it for a neater solution with fewer oversights
 
  • #10
Ray Vickson said:
So, there are a total of 5 solutions; you have given only three.

Zander Forrester said:
Thank you again.
Were you able to find the other two solutions?
 
  • #11
Thanks for your reply.
The textbook gave the three answers I submitted.
With sin(2x) = 0
2x = 0, 180, 360, 540 and 720
X = 0,90,180, 270 and 360
X = 0, pi/2, pi, 3pi/2 and 2pi.
Zander
 
  • #12
Zander Forrester said:
X = 0, pi/2, pi, 3pi/2 and 2pi.
These look like five solutions to me.
 

1. What are stationary points?

Stationary points, also known as critical points, are points on a graph where the slope is equal to zero. This means that the tangent line to the curve at that point is horizontal and the graph changes from increasing to decreasing or vice versa.

2. How do you find stationary points on a graph?

To find stationary points on a graph, you need to take the derivative of the function and set it equal to zero. Then, solve for the variable to find the x-values of the stationary points. These x-values can then be substituted into the original function to find the corresponding y-values.

3. What is the derivative of y = 3sin^2x?

The derivative of y = 3sin^2x is y' = 6sinx*cosx. This can be found using the chain rule, where the derivative of sin^2x is 2sinx*cosx.

4. How many stationary points does y = 3sin^2x have?

In general, a trigonometric function will have two stationary points per period. Since the period of sin^2x is π, the function y = 3sin^2x will have two stationary points for every π increment on the x-axis. Therefore, the number of stationary points on this graph is infinite.

5. Can stationary points exist on a flat line?

Yes, stationary points can exist on a flat line. This occurs when the slope of the line is zero, meaning the line is neither increasing nor decreasing. In this case, the stationary points would be the entire line.

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