Proof without calculus: slope of parabola

This is an image :D This is an image :D

For anyone who has learnt Calculus must know that the slope at any point on a parabola, for instance y=12x2 y = \frac{1}{2} x^2, is always equivalent to dydx=x \frac{dy}{dx} = x . Yet I was just wondering if we can prove it without Calculus? Maybe with only geometry, vector, I don't know. If you would like to try this proof, I totally appreciate your effort, since I can't complete it on my own. Share your ideas here and let me know your wonderful thoughts, thanks! :D

For anyone who is curious to take a peek at this animation I made in Desmos(regarding this problem), feel free to check it out on this link: https://www.desmos.com/calculator/qw71a3cyyl

Note by Horace Hung
1 month ago

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The slope is equal to rise over run (ΔyΔx) \Big( \frac{\Delta y}{\Delta x } \Big) . Consider two points separated by a very small xx distance ϵ \epsilon .

(x1,y1)=(x0,12x02)(x2,y2)=(x0+ϵ,12(x0+ϵ)2) (x_1, y_1) = \big(x_0, \frac{1}{2} x_0^2 \Big) \\ (x_2, y_2) = \big(x_0 + \epsilon, \frac{1}{2} (x_0 + \epsilon)^2 \Big)

The slope is the change in y y over the change in x x .

slope=y2y1x2x1=12(x0+ϵ)212x02(x0+ϵ)x0\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{1}{2} (x_0 + \epsilon)^2 - \frac{1}{2} x_0^2 }{(x_0 + \epsilon) - x_0}

Expanding gives:

slope=12x02+x0ϵ+12ϵ212x02ϵ\text{slope} = \frac{\frac{1}{2} x_0^2 + x_0 \epsilon + \frac{1}{2} \epsilon^2 - \frac{1}{2} x_0^2 }{\epsilon}

Suppose ϵ \epsilon is much smaller than x0 x_0 , meaning that the ϵ2 \epsilon^2 term in the numerator can be neglected. The expression then reduces to:

slope=x0\text{slope} = x_0

Steven Chase - 1 month ago

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That looks suspiciously like a derivative to me! :D

David Stiff - 1 month ago

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Isn't the very concept of "slope" the same thing as a derivative?

Steven Chase - 1 month ago

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@Steven Chase True. :)

I think that's what the original question is referring to though: is there a way to define/calculate slope in, say, a purely geometrical way? @Avinash Rathore seems to have described something like that.

David Stiff - 1 month ago

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@David Stiff Finally someone understands my question!!! @Steven Chase solution is in fact first principle in disguise, which I'd regard as part of Calculus :( Still a nice try. @Avinash Rathore solution is really similar to my original geometric proof, congrats!

Frankly I'd like to post my merely geometric proof here (without the slightest use of Calculus / Algebra / point-line distance formula), but the comment box is too small so it lefts as an exercise to readers

P.S. Just kidding, I'll try to post it somehow (Hint: @Avinash Rathore solution but with purely congruent triangles/locus of parabola/concepts of focus + directrix etc.)

Horace Hung - 1 month ago

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@Horace Hung Awesome! :D

David Stiff - 1 month ago

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The tangent and the normal at a point are the two angle bisectors of the lines (i) joining the focus and the point, and (ii) a line parallel to the y-axis passing through the point. Using the formula for the angle bisectors of a couple of lines,

          (Ax + By + C)^2/ (A^2 + B^2) = (ax + by + c)^2/ (a^2 + b^2)

(whose 2 roots give the tangent and normal), the line with the negative y intercept is the tangent. Simplifying, we can easily get the slope of the tangent as m = x.

NOTE - The above formula can be easily derived by equating the distance of any point on an angle bisector from the 2 given lines. This derivation does not require calculus.

Also, we note the above fact of the tangent and normal being angle bisectors by remembering the parabolic property that all lines parallel to the axis of the parabola get reflected or focused on to the focus which makes the normal at the point an angle bisector (and the tangent, being perpendicular to the normal is the other angle bisector).

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Appreciate your insight !

Horace Hung - 1 month ago

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