The point (1,0) lies on the curve y=sin(10π/x).

Question

A) if Q is the point (x,sin(10π/x), find the slope of the secant line PQ.

Points are 2,1.5,1.4,1.3,1.2,1.1,0.5,0.6,0.7,0.8,0.9

Do slopes appear to be approaching a limit?

There is no 10π/x on the unit circle..... So would you put it in this format?
2, 10π/2?
For point 2 the y equals to 1.7321? How does this happen?

Answer 1

don't use the graph of the unit circle

instead look at the graph of the function sin((10ð/x)

(1,0) lies on the curve because sin((10ð) = 0

for x = 2, y = sin(5ð), which is 0 NOT 1.7321

Answer 2

you want the points

(2, sin10π/2),(1.5, sin10π/1.5) and so on

Do not expect the slope of the secant to approach a limit. It will oscillate between values approaching -1 and 1

Answer 3

No the answer is 1.7321 if x = 2 then y is 1.7321 that is the answer in the back

Answer 4

The sin function range is between -1 and +1

1.7321 is not the sin of any angle.

Answer 5

sin(10π/2) = sin(5π) = 0
the slope of the line joining (1,0) and (2,0) is 0

something is wrong with this picture.

Answer 6

The slope of the line from the point (1.5,0.866) to the point (2,0) is -1.7321.

But the value of the function y=sin(10Pi/x) at x=2 is y=0.

The sine function will always return 0 when given an integer number times Pi, and 10/2 is an integer.