Finding Where the slope is Undefined

Question

So I was helping a student with a problem today. The answer we tried to submit was wrong and I'm not sure I understand why.

The question was:

Find where the slope is undefined for x = 4sec(y)

At first I thought this was fairly straightforward:

We implicitly differentiated, getting:

1 = 4y'sec(y)tan(y)

Then solved for y':

y' = 1/(4y'sec(y)tan(y))

Now I assumed that the slope should be undefined wherever the denominator was zero:

4sec(y)tan(y) = 0

sec(y)tan(y) = 0

Both sec(y) and tan(y) should be zero for the same values of y, namely:

y = pi/2 + k*pi

But this was the wrong answer! I'm worried it might have something to do with the fact that we're dealing with y instead of x, but I was stumped as to how to proceed.

Answer 1

Your mistake is in concluding "Both sec(y) and tan(y) should be zero for the same values of y." This is not true.

sec(y)tan(y) = 0 whenever either sec(y)=0 or tan(y)=0. Moreover, the secant function is never 0. So sec(y)tan(y) = 0 precisely when tan(y)=0, that is, when y = nπ for any integer n. Note also that the corresponding values of x are ±4.

Answer 2

If x= sec(y)=1/cos(y) then x is undefined when y is an odd multiple of π/2:

so the domain (the set of all values the independent variable can assume) is all real values except nπ/2 where n is any odd integer.

The range is all real numbers.

Graphically, this means there are n horizontal asymptotes at nπ/2 where n is an odd integer. Does this help?