Slopes and Infinity

Alright, this might seem stupid, but I never have received an answer for this question. The slope of a vertical line is undefined. Now my question is why can't the slope of the vertical line be infinity? What is the difference? How does the line with infinity slope look like?

 

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The slope of a line is the rise divided by the run. With a verticle line, the run (horozontal distance) is zero. Mathematics does not allow division by zero beause you cannot get from the answer back to your original starting place.

Ten divided by two is five. With this result, I can back into the starting point of ten by multiplying five by two. This does not work with division by zero though. Ten divided by zero is infinity. When I try to back into my original starting point, I multiply zero by infinity and come up with zero, not ten.

Because division by zero does not allow the math to run in both directions, we say that it is undefined.

 

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No... you come up with "undefined".

I see no problem with describing a vertical line as having infinite slope, unless you try to do something silly like use the y=mx+c form for such a line.

It is interesting that a line with gradient of infinity looks the same as a line with gradient of negative infinity.

 

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Me neither, except for the fact that I do not like to be wrong.

SLOPE = RISE / RUN

The RUN of a vertical line is ZERO. ANYTHING divided by zero is undefined. Not infinity. A slope that approaches vertical approaches infinity, but as soon as it is vertical, it no longer has a slope. Therefore it is undefined.

 

It wouldn't be an infinite rise/drop because the run is 0. It would be infinity if m was denoted as infinity (not x/0), however when there is a certain rise over zero run, it is undefined because the function still rises according to m however zero distance is travelled and appears to be infinite. The undefined techincally comes from the idea that you don't know what m would be, if the distance travelled was > 0.

Basically I summed up Maddad's point in a different way.

Also, infinity x zero is zero. Because multiplying anything by zero is as if you're not multiplying anything at all, bringing you back to your starting point, at nothing.

 

This should be in the math section, but I thought I'd ask since we're on the topic:

Calculus defines slope at anytime of any equation as the derivative at that point in the equation. So what's the slope at any point of the equation x=sin(y)/y?
(And unfortunately I have no idea where to begin guessing).

 

I'll just start by pointing out that the derivative is not defined where the graph is not continuous - here at y=0.
For the actual differentiation, see:
http://en.wikipedia.org/wiki/Quotient_rule

 

Actually, the prof recently did a long proof in math class where y=sinx/x at x = 0 is actually equal to 1. And that is the real answer too, because at any point when you deal with limits and get a 0/0, you can find something to cancel out and give you a non-0/0 answer.

 

Approach it from the other direction.
Is a line of infinite slope (a line with infinite rise for any run) vertical?

 

Strictly speaking, the limit of sinx/x as x approaches zero equals 1... but the value actually at zero is undefined. The function is discontinuous at that point.

 

I hope what your professor actually said was that IF you define
f(x)= (sin x)/x for x NOT 0, 1 for x= 0 THEN f is continuous.

The function f(x)= (sin x)/x is not defined at x= 0 and has a "removable discontinuity" at x= 0.

 

a vertical gradient is deffined as infinity in many places. a slope with a horizontal gradient is 0 ... which does not agree with the law for perpendicular lines that m1 x m2 = -1 , or is infinity multiplied by zero -1????

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