- Level: International Baccalaureate
- Subject: Maths
- Word count: 4205
Derivative of Sine Functions
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Introduction
Derivative of Sine Functions
Question 1
Investigate the derivative of the functionf(x) =sinx.
- Graph the function f(x) =sinx. For－2x2
ｙ＝f(x). Let f(x)=sinx ,－2x2
By sketching the graph we get:
Figure 1 : graph of function f(x) =sinx
Figure 1 reveals the range of the function f(x) =sinx. is [－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
－2 to －,－ to －，０to ， to .
The line of the tangent become more and more precipitous as the points move from left to right within － to －，－ to 0， to ， to 2.
·In the domain of : ［－2，－［， ］－,［，］,2］the gradient is positive.
The gradient is negative in the domain of:：］－，－［, ］， ［.
·At the point when x equals to －,－,and ,the gradient is obviously 0.
c) Use your Graphics Calculator (GC) to find numerical values of the gradient of the function at every /4 unit. Sketch your findings on a graph.
- The numerical values (in 3 significant figures) of the gradient of the function at every /4 unit is shown in the table below:
X | －2 | － | － | － | － | － | － | － | 0 |
1.00 | 0.707 | 0 | －0.707 | －1.00 | －0.707 | 0 | 0.707 | 1.00 | |
x | 2 | ||||||||
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 /4 unit is shown below.
Gradient of f(x) =sinx.at every /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.
Middle
］－1, 2］. The gradient is negative in the domain of:：］－－1，－－1［,
］－1， －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
－2+1 to －+1,－+1 to －+1，1to +1，+1 to +1.
The line of the tangent becomes steeper and steeper as the points move from left to right within－2to－2+1,－+1to－+1，－+1 to 1，+1 to +1，+1to 2.
·The gradient is positive in the domain of :［－2，－+1［， ］－+1, +1［，
］+1,2］
The gradient is negative in the domain of:：］－+1，－+1［, ］+1, +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
－－２to －－２，－２to－２，－２ to －２，2－２ to －２.
The line of the tangent becomes steeper and steeper as the points move from left to right within－2to －－２，－－２to－２，－２to－２, －２to 2－２, －２ to 2.
·The gradient is positive in the domain of :］－－２,－２［，］－２, －２］
The gradient is negative in the domain of:］－2,－－２［, ］－２,－２［
］－２, 2].
3. Values of gradient of the function at every /4 unit, sketch findings on a graph.
①The value of the gradient of f(x) =sin(x+1) at every /4 unit is shown in the table below:
X | －2 | － | － | － | － | － | － | － | 0 |
0.540 | -0.213 | -0.841 | －0.977 | －0.540 | 0.213 | 0.841 | 0.977 | 0.540 | |
X | 2 | ||||||||
-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)
gradient of f(x) =sin(x+1) at every /4 unit
②The value the gradient of f(x) =sin(x－1) at every /4 unit is shown in the table below:
X | －2 | － | － | － | － | － | － | － | 0 |
0.540 | 0.977 | 0.841 | 0.213 | －0.540 | －0.977 | －0.841 | －0.213 | 0.540 | |
X | 2 | ||||||||
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)
gradient of f(x) =sin(x-1) at every /4
③The value of the gradient of f(x) =sin(x+2) at every /4 unit is shown in the table below:
X | －2 | － | － | － | － | － | －...read more. Conclusion
,－,－,－,0,, , and 2, the gradient is obviously 0. ...read more.
c) Find values of gradient of the function at /4 every unit and sketch findings on a graph The numerical values of the gradient of the function at every /4 unit is showed in the table:
The scatter plot for the gradient of f(x) =sinx. at every /4 unit is shown below. gradient of f(x) =sin(x+1) at every /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)=sinx found by the calculator with the curve of m′(x)=sin2x is shown in figure 13: Figure 12 m′(x)=sin2x figure 13 derivative of m(x)=sinx 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)=sinx. 2. Method and findings: The curve in figure 16 reveals the following characteristics: ·symmetrical along the axis of . ·repeat its values every ,(so the function is periodic with a period of ) ·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 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)=sinx is m′(x)=sin2x and it can be written as m′(x)=2sinxcosx. This student written piece of work is one of many that can be found in our International Baccalaureate Maths section. Found what you're looking for?
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