# Thread: Total Derivative and Orthogonality

1. ## Total Derivative and Orthogonality

I did the first part by the definition of total derivative, but can not think of the way to do the second part of the question. Any help please?

2. If $f(x)\in\partial B_r(0)$ then $(f_1(x))^2 + (f_2(x))^2 + \ldots + (f_n(x))^2 = r^2$ for all x in U. Differentiate that equation.

3. Originally Posted by Opalg
If $f(x)\in\partial B_r(0)$ then $(f_1(x))^2 + (f_2(x))^2 + \ldots + (f_n(x))^2 = r^2$ for all x in U. Differentiate that equation.
Thanks for the very quick response.
Could you please elaborate what you meant by differentiate the equation?
and how would that help show a vector is orthogonal to (DF)a?

4. Originally Posted by 6DOM
Thanks for the very quick response.
Could you please elaborate what you meant by differentiate the equation?
and how would that help show a vector is orthogonal to (DF)a?
I meant differentiate with respect to x, using the chain rule. I suppose I should have used t rather than x, since the question says that f is a function of t. Then $\tfrac d{dt}(f_1(t))^2$ evaluated at t=a is equal to $2f_1(a)\tfrac {df_1}{dt}(a)$, and similarly for the other coordinates. The result of differentiating both sides of the equation $(f_1(t))^2 + (f_2(t))^2 + \ldots + (f_n(t))^2 = r^2$ in that way ought to remind you of an inner product.

5. Originally Posted by Opalg
I meant differentiate with respect to x, using the chain rule. I suppose I should have used t rather than x, since the question says that f is a function of t. Then $\tfrac d{dt}(f_1(t))^2$ evaluated at t=a is equal to $2f_1(a)\tfrac {df_1}{dt}(a)$, and similarly for the other coordinates. The result of differentiating both sides of the equation $(f_1(t))^2 + (f_2(t))^2 + \ldots + (f_n(t))^2 = r^2$ in that way ought to remind you of an inner product.
Understand it perfectly now. Thanks!