# Local Extreme Values Question

• Nov 1st 2008, 10:13 AM
sazafraz
Local Extreme Values Question
Let y=f(x) be differentiable and suppose that the graph of f does not pass through the origin. The distance D from the origin to a point P(x,f(x)) of the graph is given by

$D=\sqrt{x^2 + [f(x)]^2}$

Show that if D has a local extreme value at c, then the line through (0,0) and (c,f(c)) is perpendicular to the line tangent to the graph of f at (c,f(c))

Thanks
• Nov 1st 2008, 06:19 PM
mr fantastic
Quote:

Originally Posted by sazafraz
Let y=f(x) be differentiable and suppose that the graph of f does not pass through the origin. The distance D from the origin to a point P(x,f(x)) of the graph is given by

$D=\sqrt{x^2 + [f(x)]^2}$

Show that if D has a local extreme value at c, then the line through (0,0) and (c,f(c)) is perpendicular to the line tangent to the graph of f at (c,f(c))

Thanks

Don't bump.

You should realise that the gradient of the tangent is $f'(c)$ and the gradient of the line is $\frac{f(c)}{c}$. If they are normal then their product is equal to -1: $f'(c) \cdot \frac{f(c)}{c} = -1 \Rightarrow f'(c) \cdot f(c) = -c$.

Use the chain rule to differentiate D:

$\frac{d D}{dx} = \frac{1}{2 \sqrt{x^2 + [f(x)]^2}} \cdot [2x + 2 f(x) f'(x)]$.

$\frac{d D}{dx} = 0 \Rightarrow 2x + 2 f(x) f'(x) = 0 \Rightarrow f(x) \, f'(x) = -x$.

Therefore ....