raxAdaam
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Hi -
I was reading Adams' "Calculus: A Complete Course" (6th edition) and he offers the following proof that the gradient of a function:
I'm just wondering why the emphasis on specifying the values of x(0) and y(0), or even the need to specify t = 0 in the last statement? If \mathbf{r} is a parameterization of the level curve, then isn't the derivative of f zero for all values of t? Why not go for the more general statement?
I guess I just don't find the proof very satisfying. Obviously I don't doubt it, but does anyone know a more intuitive proof? I'm teaching the concept shortly and always appreciate bringing an intuitive approach. Usually I try to make connections to single variable, but that's not possible here …
Thanks in advance,
Rax
I was reading Adams' "Calculus: A Complete Course" (6th edition) and he offers the following proof that the gradient of a function:
Let \mathbf{r} = x(t)\mathbf{i} + y(t)\mathbf{j} be a parametriazation of the level curve of f such that x(0) = a and y(0) = b. Then for all t near 0, f\left( x(t), y(t)\right) = f(a,b). Differentiating this equation with respect to t using the chain rule, we obtain:
f_1\frac{dx}{dt}+ f_2\frac{dy}{dt} = 0
At t=0 this says that \nabla f(a,b) \cdot \frac{d\mathbf{r}}{dt}\Big|_{t=0}=0, that is, \nabla f is perpendicular to the tangent vector \frac{d\mathbf{r}}{dt} to the level curve at (a,b).
I'm just wondering why the emphasis on specifying the values of x(0) and y(0), or even the need to specify t = 0 in the last statement? If \mathbf{r} is a parameterization of the level curve, then isn't the derivative of f zero for all values of t? Why not go for the more general statement?
I guess I just don't find the proof very satisfying. Obviously I don't doubt it, but does anyone know a more intuitive proof? I'm teaching the concept shortly and always appreciate bringing an intuitive approach. Usually I try to make connections to single variable, but that's not possible here …
Thanks in advance,
Rax