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I'm linking the following proof from MIT OCW about why the gradient vector is normal to the surface.

https://ocw.mit.edu/courses/18-02sc-multivariable-calculus-fall-2010/85c1d85363d9808505351b571d2750aa_MIT18_02SC_notes_19.pdf

Now, while I understood the proof from the link, the first thing that came to my mind was, "Why not try and apply it to calculate derivatives of single variable functions?" However, this is when everything essentially broke for me.

Take for example, $f(x)=x^2$. Going by MIT's proof, should I not also be able to say $f(x) = x^2 = c$, as some constant $c$? From there, differentiate both sides by x, and therefore get $2x=0$? However, here is where I get confused. I simply do not know how to interpret this equation as unlike in the proof, there does not seem to be any dot product, unless of course, we were to say that 2 and x are single dimensional vectors respectively.

I suspect that I am also confusing some other concepts, and I apologize for that. I really appreciate any clarification on this.

Daanyal Ali Akhtar
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2 Answers2

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In one dimension this does not make sense as you cannot be perpendicular in any meaningful way. But to make sense of your example, consider $y = x^2$ instead. This means $y - x^2 = 0$, so we define $f(x,y) := y - x^2$ and look at the level surface $f(x,y) = 0$, which is exactly the parabola you have in mind. Then $\textrm{grad} f = \begin{pmatrix} 1 \\ -2x \end{pmatrix}$ and indeed, this vector is perpendicular to the parabola for every $x$.

Klaus
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  • Does this mean that in the same way they differentiated both sides of $g(t) = (x(t), y(t), z(t))= c$ to get the overall derivative of g with respect to t, I cannot do the same with $f(x) =x^2 = c$ to get the derivative of f with respect to x? Thanks again. – Daanyal Ali Akhtar Aug 04 '22 at 14:21
  • @DaanyalAliAkhtar The problem is that $x^2 = c$ defines at most two points. It's not a surface you can be perpendicular to. – Klaus Aug 04 '22 at 14:24
  • Thank you very much for this clarification. – Daanyal Ali Akhtar Aug 04 '22 at 14:26
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I would like to draw your attention , to something important in calculus and specially in this subject , that really helped me to overcome every question

note : u defined f(x) = x^2 = 0 , that's wrong , the correct form is f(x,y) = c (implicit function) for 2d , and not f(x)=0, and from here I will lead with Klaus's Answer

y−x^2=0 , so we define f(x,y):=y−x^2 and look at the level surface f(x,y)=0, which is exactly the parabola you have in mind. Then gradf=(1−2x) and indeed, this vector is perpendicular to the parabola for every x.

Mayyas Haj
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  • Thanks so much. However, is there anything wrong with taking the derivative of $f(x) = x^2 = c$ over $f(x,y) = y-x^2 = c$? I’m just having a hard time understanding why it must be in implicit form. Thanks. – Daanyal Ali Akhtar Aug 04 '22 at 18:13
  • You can define f(x)= x^2 , just if you define y=f(x) , and from here you can lead with y-x^2=0 , it's the same thing at last , but the Idea is you can't define a constant equal to variable , let's say u want to choose c = 0 and x = 2 , that's not possible , but if you choose y-x^2=c , x=2 , y=4 , c=0 , that's always possible , because you have 2 variables you can change – Mayyas Haj Aug 05 '22 at 11:22