Finding the Gradient of Stationary Points in Trigonometric Functions

In summary: I don't have a calculator with me to divide out pi.ok. i'll have a play and post back if i can't do it.. Thanks for the inspiration!
  • #1
pip_beard
14
0

Homework Statement


gradient of the stationary points of y=1-2sinx domain 0<x<2pi

Homework Equations


The Attempt at a Solution


dy/dx = -2cosx
-2cosx=0...?
 
Last edited:
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  • #2
Gradient of the stationary point? Do you need to find that or the gradient? Because when you put dy/dx=0 it sort of implies what the gradient will be.
 
  • #3
rock.freak667 said:
Gradient of the stationary point? Do you need to find that or the gradient? Because when you put dy/dx=0 it sort of implies what the gradient will be.

Sorry. the qu reads 'the curve y=1-2sinx has domain 0<x<pi. find the gradients of the curve at the points where the curve crosses the x-axis
 
  • #4
pip_beard said:
Sorry. the qu reads 'the curve y=1-2sinx has domain 0<x<pi. find the gradients of the curve at the points where the curve crosses the x-axis

Ah that is different.

Well you know the gradient function is dy/dx.

You need to find when the curve crosses the x-axis, or when y=0.
 
  • #5
so do i differentiate it and make it = 0?

so i get -2cosx=0
 
  • #6
pip_beard said:
so do i differentiate it and make it = 0?

so i get -2cosx=0

how to i rearrange this to work it out?
 
  • #7
pip_beard said:
so do i differentiate it and make it = 0?

so i get -2cosx=0

pip_beard said:
how to i rearrange this to work it out?

No if y=1-2sinx, the curve crosses, the x-axis when y=0

so solve 1-2sinx=0 for x. Can you do that?
 
  • #8
-2sinx=1
sinx=-1/2
 
  • #9
pip_beard said:
-2sinx=1
sinx=-1/2

which gives x as ? ( in the given domain that is)
 
  • #10
principle value of -0.523 (but out of range) the pi-(-0.523) = 3.6651

therfore value = 3.6651 and 6.2832?
 
  • #11
pip_beard said:
-2sinx=1
sinx=-1/2

pip_beard said:
principle value of -0.523 (but out of range) the pi-(-0.523) = 3.6651

therfore value = 3.6651 and 6.2832?



this should be sinx=1/2.


But how did you get that value for x?
 
  • #12
ohhh... did sin^-1... so ...

x principle = 0.479425
so pi - 0.4879425 = 2.66216
 
  • #13
pip_beard said:
ohhh... did sin^-1... so ...

x principle = 0.479425
so pi - 0.4879425 = 2.66216

you should get x= π/6 and another one, best to use the exact values, I don't have a calculator with me to divide out pi.
 
  • #14
ok. i'll have a play and post back if i can't do it.. Thanks for the inspiration!
 

1. What is the gradient of a stationary point?

The gradient of a stationary point is the slope or rate of change of a function at that particular point. It indicates the direction and steepness of the function at the point where it is neither increasing nor decreasing.

2. How do you calculate the gradient of a stationary point?

To calculate the gradient of a stationary point, you need to take the derivative of the function at that point. This can be done by using the derivative rules such as the power rule, product rule, or chain rule depending on the form of the function.

3. What does a positive gradient at a stationary point indicate?

A positive gradient at a stationary point indicates that the function is increasing at that point. This means that as you move along the function in the positive direction, the function values will also increase.

4. How do you determine if a stationary point is a maximum or minimum?

To determine if a stationary point is a maximum or minimum, you can use the second derivative test. If the second derivative is positive, then the stationary point is a minimum. If the second derivative is negative, then the stationary point is a maximum. If the second derivative is zero, then the test is inconclusive.

5. Can a stationary point have a zero gradient?

Yes, a stationary point can have a zero gradient. This means that the function has a horizontal tangent at that point, and the function values are neither increasing nor decreasing at that point. In this case, the stationary point can be a maximum, minimum, or a point of inflection.

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