- #1
Rasalhague
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Spivak: Calculus on Manifolds, p. 42:
Assuming "differentiating [itex]f(x,g(x)) = 0[/itex]" means differentiating [itex]f \circ h[/itex], where
[tex]h : (-1,1) \rightarrow \mathbb{R}^2 \; | \; x \mapsto (I(x),g(x)), I(x) = x[/tex]
and setting the result identically equal to zero, we have
[tex]2x+2g(x) \cdot g'(x) = 0,[/tex]
as above. Substituting
[tex]g(x) = \sqrt{1-x^2},[/tex]
I get
[tex]=2x-\frac{2 \sqrt{1-x^2} \cdot 2x}{\sqrt{1-x^2}}[/tex]
[tex]=-2x = 0.[/tex]
Or, if I substitute the negative square root, [itex]6x = 0[/itex]. Therefore, either way, [itex]x = 0[/itex]. But hang on! By definition, [itex]g[/itex] maps (-1,1) to the reals, its graph forming a semicircle; how can this other equation be telling us that [itex]x=0[/itex]?
Reconsider the function [itex]f:\mathbb{R}^2 \rightarrow \mathbb{R}[/itex] defined by [itex]f(x,y) = x^2 + y^2 - 1[/itex], we note that the two possible functions satisfying [itex]f(x,g(x)) = 0[/itex] are
[tex]g(x) = \sqrt{1-x^2}[/tex]
and
[tex]g(x) = -\sqrt{1-x^2}.[/tex]
Differentiating [itex]f(x,g(x)) = 0[/itex] gives
[tex]D_1(x,g(x))+D_2(x,g(x)) \cdot g'(x) = 0[/tex]
or
[tex]2x +2g(x) \cdot g'(x) = 0,[/tex]
[tex]g'(x) = -x/g(x),[/tex]
which is indeed the case for either
[tex]g(x) = \sqrt{1-x^2}[/tex]
or
[tex]g(x) = -\sqrt{1-x^2}[/tex].
Assuming "differentiating [itex]f(x,g(x)) = 0[/itex]" means differentiating [itex]f \circ h[/itex], where
[tex]h : (-1,1) \rightarrow \mathbb{R}^2 \; | \; x \mapsto (I(x),g(x)), I(x) = x[/tex]
and setting the result identically equal to zero, we have
[tex]2x+2g(x) \cdot g'(x) = 0,[/tex]
as above. Substituting
[tex]g(x) = \sqrt{1-x^2},[/tex]
I get
[tex]=2x-\frac{2 \sqrt{1-x^2} \cdot 2x}{\sqrt{1-x^2}}[/tex]
[tex]=-2x = 0.[/tex]
Or, if I substitute the negative square root, [itex]6x = 0[/itex]. Therefore, either way, [itex]x = 0[/itex]. But hang on! By definition, [itex]g[/itex] maps (-1,1) to the reals, its graph forming a semicircle; how can this other equation be telling us that [itex]x=0[/itex]?