Question

(a) If Q is the point (x, sin(16)),(1) 2(11)15Xfind the slope of the secant line PQ (correct to four decimal places) for the following values of x.

Answer 1

Given:

The point P is (1, 0).

The point Q is

`Q(x,sin((16\pi)/(x)))`

Required:

We need to find the slope of the scent line PQ when x=2.

Step-by-step explanation:

i)

Replace x =2 in point Q.

`Q(2,sin((16\pi)/(2)))`
`Q(2,sin(8\pi))`

Consider the slope formula.

`m=(y_2-y_1)/(x_2-x_1)`
`Substitute\text{ }x_1=1,y_1=0,x_2=2,\text{ and }y_2=sin(8\pi)\text{ in the formula.}`
`m=(sin(8\pi)-0)/(2-1)`
`Use\text{ sin\lparen}8\pi)=0.`
`m=(0)/(1)=0`

The slope when x=2 is 0.0000.

ii)

Replace x =1.5 in point Q.

`Q(1.5,sin((16\pi)/(1.5)))`
`Q(1.5,sin((32\pi)/(3)))`

Consider the slope formula.

`m=(y_2-y_1)/(x_2-x_1)`
`Subst\imaginaryI tute\text{ x}_1=1,y_1=0,x_2=1.5,\text{ and }y_2=s\imaginaryI n((32\pi)/(3))\text{ }\imaginaryI\text{n the formula}`
`m=(sin((32)/(3)\pi)-0)/(1.5-1)`
`m=1.7320508`
`m=1.7321`

The slope when x=1.5 is 1.7321.

iii)

Replace x =1.4 in point Q.

`Q(1.4,sin((16\pi)/(1.4)))`
`Q(1.4,sin((80\pi)/(7)))`

Consider the slope formula.

`m=(y_2-y_1)/(x_2-x_1)`
`Subst\imaginaryI tute\text{ x}_1=1,y_1=0,x_2=1.4,\text{ and }y_2=s\imaginaryI n((80\pi)/(7))\text{ }\imaginaryI\text{n the formula}`
`m=(sin((80)/(7)\pi)-0)/(1.4-1)`
`m=-2.43731978045`
`m=-2.4373`

The slope when x=1.4 is -2.4373.

iv)

Replace x =1.3 in point Q.

`Q(1.3,sin((16\pi)/(1.3)))`
`Q(1.3,sin((160\pi)/(13)))`

Consider the slope formula.

`m=(y_2-y_1)/(x_2-x_1)`
`Subst\imaginaryI tute\text{ x}_1=1,y_1=0,x_2=1.3,\text{ and }y_2=s\imaginaryI n((160\pi)/(13))\text{ }\imaginaryI\text{n the formula}`
`m=(sin((160)/(13)\pi)-0)/(1.3-1)`
`m=2.74327955298`
`m=2.7433`

The slope when x=1.3 is 2.7433.

v)

Replace x =1.2 in point Q.

`Q(1.2,sin((16\pi)/(1.2)))`

Consider the slope formula.

`m=(y_2-y_1)/(x_2-x_1)`
`Subst\imaginaryI tute\text{ x}_1=1,y_1=0,x_2=1.2,\text{ and }y_2=s\imaginaryI n((16\pi)/(1.2))\text{ }\imaginaryI\text{n the formula}`
`m=(sin((16)/(1.2)\pi)-0)/(1.2-1)`
`m=-4.33012701892`
`m=-4.3301`

The slope when x=1.2 is -4.3301.

vi)

Replace x =1.1 in point Q.

`Q(1.1,sin((16\pi)/(1.1)))`

Consider the slope formula.

`m=(y_2-y_1)/(x_2-x_1)`
`Subst\imaginaryI tute\text{ x}_1=1,y_1=0,x_2=1.1,\text{ and }y_2=s\imaginaryI n((16\pi)/(1.1))\text{ }\imaginaryI\text{n the formula}`
`m=(sin((16)/(1.1)\pi)-0)/(1.1-1)`
`m=9.89821441881`
`m=9.8982`

The slope when x=1.1 is 9.8982

vii)

Replace x =0.5 in point Q.

`Q(0.5,sin((16\pi)/(0.5)))`

Consider the slope formula.

`m=(y_2-y_1)/(x_2-x_1)`
`Subst\imaginaryI tute\text{ x}_1=1,y_1=0,x_2=0.5,\text{ and }y_2=s\imaginaryI n((16\pi)/(0.5))\text{ }\imaginaryI\text{n the formula}`
`m=(sin((16)/(0.5)\pi)-0)/(0.5-1)`
`m=0`
`m=0.0000`

The slope when x=0.5 is 0.0000.

viii)

Replace x =0.6 in point Q.

`Q(0.6,sin((16\pi)/(0.6)))`

Consider the slope formula.

`m=(y_2-y_1)/(x_2-x_1)`
`Subst\imaginaryI tute\text{ x}_1=1,y_1=0,x_2=0.6,\text{ and }y_2=s\imaginaryI n((16\pi)/(0.6))\text{ }\imaginaryI\text{n the formula}`
`m=(sin((16)/(0.6)\pi)-0)/(0.6-1)`
`m=-2.16506350946`
`m=-2.1651`

The slope when x=0.6 is -2.1651

Similarly, we get

The slope when x=0.7 is -1.4463

The slope when x=0.8 is 0.000

The slope when x=0.9 is 6.4279

We see that the values are not approaching any particular value.

As x approaches 1, the slope does not appear to be approaching any particular value.

B)

The graph of the given curve is

There are frequency oscillations near point 1.

We need to take x-values further to 1 to get the exact slope.

C)

The given curve is

`f(x)=sin((16\pi)/(x))`
`Slope\text{ of the secent line=}(f(a+h)-f(a))/(h)`

Here a=1.

`Slope\text{ of the secent line=}(f(1+h)-f(1))/(h)`
`Slope\text{ of the secent line=}(sin((16\pi)/(1+h))-sin(16\pi))/(h)`
`Slope\text{ of the secent line=}(sin((16\pi)/(1+h)))/(h)`

Take limit h tends to zero.

Differentiate with respect to h.