# Total Derivative and Orthogonality

• Oct 25th 2009, 06:35 AM
6DOM
Total Derivative and Orthogonality
http://i33.tinypic.com/2wce634.png

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?
• Oct 25th 2009, 08:29 AM
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.
• Oct 25th 2009, 09:24 AM
6DOM
Quote:

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?
• Oct 25th 2009, 12:10 PM
Opalg
Quote:

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.
• Oct 25th 2009, 12:22 PM
6DOM
Quote:

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!