Find the slope of secant line PQ and find the tangent line at point P(1,0) for function y = sin (2pi/x)

Question

x ---> 1-

I am not sure what numbers to use to approach the slope of PQ from the left and right.

I have tried pi/8, pi/6, pi/5, pi/4 for numbers approaching 1 from the left.

x ---> 1+

For numbers that approach 1 from the right, I have tried 3pi/2, pi, 2pi/3, pi/2, and pi/3.

I know that the slope is supposed to be -2pi, but I am unsure how to solve it.

We haven't learned to find the derivative as the semester just started or use limits.

Thank you in advance!

Answer 1

I'm not sure what limiting you have learned, but the secant (let's say from below) will have a slope Δy/Δx

Let's pick the slope from x = 1-h to 1: slope of secant = sin(2π)-sin(2π/(1-h))/(1-(1-h)) = 1/h*(sin(2π)-sin(2π/(1-h)))

If we take the limit as h goes to zero to get at the secant approaching the tangent, plugging in doesn't work as we have (0-0)/0

You may know the binomial theorem: (1+ε)ⁿ ∼ 1+nε if ε<< 1 for all real n. In this case we have h<<1 and 1/(1-h) or (1-h)^-1 approaches 1-(-h) or 1+h. So...

Tangent slope = Lim as h→0 1/h (sin(2π) - sin(2π(1+h))

We can use the sum of angles sin formula (second term is sin(2π+2πh)

lim as h →0 1/h(sin(2π) - sin(2π)cos(2πh) - cos(2π)sin(2πh)) Because cos(2πh) approaches 1 as h approaches 0, the first two terms cancel leaving 1/h (-cos(2π)(sin(2πh))).

cos(2π) = 1 so we are left with the limit as h approaches 0 of -sin(2πh)/h

Often, when you do limits, you learn that lim x->0 of sinx is x.

which results in -2pih/h = -2pi

Now you just need to write the equation of a line with slope -2π that goes through (1,0)

You may have other tricks available to figure out the limiting value of the secant.

Another, numerical way to do it is to draw a secant that is above and below and solve for the slope.

sin(2pi/(1+h)) - sin(2pi/(1-h)) /2h for h is .1, .01, .001 and you will see that it approaches -6.28.