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Derivative of Sine Functions

Extracts from this document...

Introduction

Derivative of Sine Functions

Question 1

Investigate the derivative of the functionf(x) =sinx.

  1. Graph the function f(x) =sinx. For-2image10.pngimage11.pngximage11.png2image10.png

y=f(x). Let f(x)=sinx   ,-2image10.pngimage11.pngximage11.png2image10.png

  By sketching the graph we get:image00.png

image49.png Figure 1 : graph of function f(x) =sinximage08.pngimage06.pngimage07.pngimage04.pngimage05.pngimage02.pngimage03.pngimage01.png

Figure 1 reveals the range of the function f(x) =sinx. is [-1,1].

  1. Based on this graph, describe as carefully and fully as you can, the behabiour of the gradient of the function on the given domain.

The gradient of each point on the curve is valued as the gradient of the tangent line of the point.

According to the curve in figure 1, the behaviour of the gradient of the function indicates the following characteristics:

·The line of the tangent become flatter and flatter as the points move from left to right within

-2image10.png to -image12.pngimage10.png,-image10.png to -image18.pngimage16.png,0to image18.pngimage10.pngimage10.png to image12.pngimage10.png.

The line of the tangent become more and more precipitous as the points move from left to right within -image12.pngimage10.png to -image10.png,-image18.pngimage16.png to 0,image18.pngimage10.png to image10.pngimage12.pngimage10.png to 2image10.png.

·In the domain of : [-2image10.png,-image12.pngimage10.png[, ]-image18.pngimage16.png,image18.pngimage10.png[,]image12.pngimage10.png,2image10.png]the gradient is positive.

The gradient is negative in the domain of::]-image12.pngimage10.png,-image18.pngimage16.png[, ]image18.pngimage10.pngimage12.pngimage10.png[.

·At the point when x equals to -image12.pngimage10.png,-image18.pngimage16.png,image18.pngimage10.pngand image12.pngimage10.png,the gradient is obviously 0.

c) Use your Graphics Calculator (GC) to find numerical values of the gradient of the function at every image10.png/4 unit. Sketch your findings on a graph.

  1. The numerical values (in 3 significant figures) of the gradient of the function at every image10.png/4 unit is shown in the table below:

X

-2image13.png

image14.pngimage10.png

image12.pngimage10.png

image15.pngimage16.png

image10.png

image17.pngimage10.png

image18.pngimage16.png

image19.pngimage10.png

0

image20.png

1.00

0.707

0

-0.707

-1.00

-0.707

0

0.707

1.00

x

image19.pngimage10.png

image18.pngimage16.png

image17.pngimage10.png

image10.png

image15.pngimage16.png

image12.pngimage10.png

image14.pngimage10.png

2image10.png

image20.png

0.707

0

-0.707

-1.00

-0.707

0

0.707

1.00

2. The scatter plot (joined by a smooth curve) for the gradient of f(x) =sinx. at every image10.png/4 unit is shown below.

image53.png Gradient of f(x) =sinx.at every image10.png/4 unit

d) make a conjecture for the derived function f(x)

conjecture: f(x) = cosx

e)Use your GC to test your conjecture graphically. Explain your method and your findings. Modify your conjecture if necessary

1.

...read more.

Middle

image12.pngimage10.png-1[,]-image18.pngimage16.png-1, image18.pngimage10.png-1[,

image12.pngimage10.png-1, 2image10.png]. The gradient is negative in the domain of::]-image12.pngimage10.png-1,-image18.pngimage16.png-1[,

image18.pngimage10.png-1, image12.pngimage10.png-1[.

② when c =-1 f(x) =sin(x-1)

·The line of the tangent becomes flatter and flatter as the points move from left to right within

-2image10.png+1 to -image12.pngimage10.png+1,-image10.png+1 to -image18.pngimage16.png+1,1to image18.pngimage10.png+1,image10.png+1 to image12.pngimage10.png+1.

The line of the tangent becomes steeper and steeper as the points move from left to right within-2image10.pngto-2image10.png+1,-image12.pngimage10.png+1to-image10.png+1,-image18.pngimage16.png+1 to 1,image18.pngimage10.png+1 to image10.png+1,image12.pngimage10.png+1to 2image10.png.

·The gradient is positive in the domain of :[-2image10.png,-image30.pngimage10.png+1[, ]-image18.pngimage16.png+1,image18.pngimage10.png +1[,

image12.pngimage10.png+1,2image10.png

The gradient is negative in the domain of::]-image12.pngimage10.png+1,-image18.pngimage16.png+1[, ]image18.pngimage10.png+1, image12.pngimage10.png+1[.

③when c =2 f(x) =sin(x+2)

·The line of the tangent becomes flatter and flatter as the points move from left to right within

image10.png-2to -image18.pngimage16.png-2,-2toimage18.pngimage10.png-2,image10.png-2 to image12.pngimage10.png-2,2image10.png-2 to image31.pngimage10.png-2.

The line of the tangent becomes steeper and steeper as the points move from left to right within-2image10.pngto -image10.png-2,-image18.pngimage16.png-2to-2,image18.pngimage10.png-2toimage10.png-2, image12.pngimage10.png-2to 2image10.png-2, image31.pngimage10.png-2 to 2image10.png.

·The gradient is positive in the domain of :]-image18.pngimage16.png-2,image18.pngimage10.png-2[,]image12.pngimage10.png-2, image31.pngimage10.png-2]

The gradient is negative in the domain of:]-2image10.png,-image18.pngimage16.png-2[, ]image18.pngimage10.png-2,image12.pngimage10.png-2[

image31.pngimage10.png-2, 2image10.png].

3. Values of gradient of the function at every image10.png/4 unit, sketch findings on a graph.

①The value of the gradient of f(x) =sin(x+1) at every image10.png/4 unit is shown in the table below:

X

-2image13.png

image14.pngimage10.png

image12.pngimage10.png

image15.pngimage16.png

image10.png

image17.pngimage10.png

image18.pngimage16.png

image19.pngimage10.png

0

image20.png

0.540

-0.213

-0.841

-0.977

-0.540

0.213

0.841

0.977

0.540

X

image19.pngimage10.png

image18.pngimage16.png

image17.pngimage10.png

image10.png

image15.pngimage16.png

image12.pngimage10.png

image14.pngimage10.png

2image10.png

image20.png

-0.213

-0.841

-0.977

-0.540

0.213

0.841

0.977

0.540

The scatter plots of values in the table is sketched below: (joined by a smooth curve)

image32.png gradient of f(x) =sin(x+1) at every image10.png/4 unit

The value the gradient of f(x) =sin(x-1) at every image10.png/4 unit is shown in the table below:

X

-2image13.png

image14.pngimage10.png

image12.pngimage10.png

image15.pngimage16.png

image10.png

image17.pngimage10.png

image18.pngimage16.png

image19.pngimage10.png

0

image20.png

0.540

0.977

0.841

0.213

-0.540

-0.977

-0.841

-0.213

0.540

X

image19.pngimage10.png

image18.pngimage16.png

image17.pngimage10.png

image10.png

image15.pngimage16.png

image12.pngimage10.png

image14.pngimage10.png

2image10.png

image20.png

0.997

0.841

0.213

-0.540

-0.977

-0.841

-0.213

0.540

The scatter plots of values in the table is sketched below: (joined by a smooth curve)

image33.png gradient of f(x) =sin(x-1) at every image10.png/4

The value of the gradient of f(x) =sin(x+2) at every image10.png/4 unit is shown in the table below:

X

-2image13.png

image14.pngimage10.png

image12.pngimage10.png

image15.pngimage16.png

image10.png

image17.pngimage10.png

image18.png...read more.

Conclusion

image13.png,-image12.pngimage10.png,-image10.png,-image18.pngimage16.png,0,image18.pngimage10.png, image10.png,image12.pngimage10.png and 2image10.png, the gradient is obviously 0.

c) Find values of gradient of the function at image10.png/4 every unit and sketch findings on a graph

The numerical values of the gradient of the function at every image10.png/4 unit is showed in the table:

X

-2image13.png

image14.pngimage10.png

image12.pngimage10.png

image15.pngimage16.png

image10.png

image17.pngimage10.png

image18.pngimage16.png

image19.pngimage10.png

0

image20.png

0

1

0

-1

0

1

0

-1

0

X

image19.pngimage50.png

image18.pngimage16.png

image17.pngimage10.png

image10.png

image15.pngimage16.png

image12.pngimage10.png

image14.pngimage10.png

2image10.png

image20.png

1

0

-1

0

1

0

-1

0

The scatter plot for the gradient of f(x) =sinx. at every image10.png/4 unit is shown below.

image51.png gradient of f(x) =sin(x+1) at every image10.png/4

d) Make a conjecture for the derive function m′(x)

the conjecture for the derived function is m′(x)=sin2x.

e) The scatter plot for the values in the table above with the function m′(x)=sin2x is shown in figure 12. The derivative of m(x)=sinimage47.pngx found by the calculator with the curve of

m′(x)=sin2x is shown in figure 13:

image52.pngimage42.png

Figure 12 m′(x)=sin2x                      figure 13 derivative of m(x)=sinimage47.pngx

Figure 12 reveals the curve of m′(x)=sin2x fits the scatter plot well. Figure 13 reveals that the derivative function found by calculator overlaps m′(x)=sin2x). Perfect.

Therefore m′(x)=sin2x is the correct conjecture for function m(x)=sinimage47.pngx.

2. Method and findings:

The curve in figure 16 reveals the following characteristics:

 ·symmetrical along the axis of image19.pngimage10.png .                

·repeat its values every image10.png,(so the function is periodic with a period of image10.png)

·the maximum is 1 the minimum is -1

·the mean value of the function is zero

·the amplitude of the function is 1

According to the characteristics the form of function for the gradient is similar to a cosine function with parameter of 1 and horizontal stretched by image18.png units.

Therefore conjecture of m′(x) is: m′(x)=sin2x

Based on the double angle formula: sin2x=2sinxcosx.

the conjecture m′(x)=sin2x,

Combine the two equation together we get: m′(x)=sin2x=2sinxcosx.

Therefore m′(x)=sin2x is the same as m′(x)= 2sinxcosx

So the derivative of functio m(x)=sinimage47.pngx is m′(x)=sin2x and it can be written as

m′(x)=2sinxcosx.

...read more.

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