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Question Number 2249 by Filup last updated on 11/Nov/15

$$\mathrm{With}\mathrm{linear}\mathrm{functions}f\left(x\right)\mathrm{and}g\left(x\right),$$$$\mathrm{if}f\left(x\right)\mathrm{\perp }g\left(x\right),\mathrm{then}:$$$$m_f{m}_g=-1\mathrm{where}{m}_i\mathrm{is}\mathrm{the}\mathrm{gradient}$$$$\mathrm{of}\mathrm{function}i\left(x\right).$$$$$$$$\mathrm{Does}\mathrm{that}\mathrm{therefore}\mathrm{mean}\mathrm{that},\mathrm{if}\mathrm{given}$$$$\mathrm{function}(\mathrm{including}\mathrm{non}-\mathrm{linear})f\left(x\right),$$$$f^{\prime }\left(x\right)g^{\prime }\left(x\right)=-1,\mathrm{if}f\left(x\right)\mathrm{\perp }g\left(x\right)\mathrm{at}x=n?$$$$$$$$e.g.$$$$f\left(x\right)=x^2$$$$f^{\prime }\left(x\right)=2x$$$$$$$$\mathrm{if}f\left(x\right)\mathrm{\perp }g\left(x\right)\mathrm{at}x=n:$$$$\therefore f^{\prime }\left(x\right)g^{\prime }\left(x\right)=-1$$$$g^{\prime }\left(x\right)=-\frac{1}{2x}$$$$g\left(x\right)=-\int \frac{1}{2x}dx$$$$\therefore g\left(x\right)=-\frac{1}{2}\ln \left(2x\right)+c_1$$$$$$$$f\left(x\right)=x^2,g\left(x\right)=-\frac{1}{2}\ln \left(2x\right)$$$$\therefore x^2(-\frac{1}{2}\ln \left(2x\right))=-1$$$$$$$$\mathrm{My}\mathrm{question}\mathrm{therefore}\mathrm{comes}\mathrm{down}\mathrm{to}:$$$$\mathrm{Are}\mathrm{the}\mathrm{tangent}\mathrm{lines}\mathrm{of}f\left(x\right)\mathrm{and}g\left(x\right)$$$$\mathrm{at}x=n\mathrm{perpendicular}\mathrm{to}\mathrm{each}\mathrm{other}$$$$\mathrm{for}\mathrm{all}\mathrm{values}\mathrm{of}n\in \{f\left(x\right),g\left(x\right)\}?$$

Answered by Filup last updated on 12/Nov/15

$$eqn\mathrm{of}\mathrm{tangent}\mathrm{line}\mathrm{on}f\left(x\right)\mathrm{at}x=n:$$$$y_f=f^{\prime }\left(n\right)(x-n)+f\left(n\right)$$$$\therefore egn\mathrm{of}\mathrm{line}\mathrm{on}g\left(x\right)\mathrm{at}x=n:$$$$f^{\prime }\left(x\right)g^{\prime }\left(x\right)=1$$$$y_g=g^{\prime }\left(n\right)(x-n)+g\left(n\right)$$$$\therefore y_g=-\frac{1}{f^{\prime }\left(x\right)}(x-n)+g\left(n\right)$$$$$$$$y_f=f^{\prime }\left(n\right)(x-n)+f\left(n\right)$$$$y_g=-\frac{1}{f^{\prime }\left(n\right)}(x-n)+g\left(n\right)$$$$$$$$y_f\mathrm{and}{y}_g\mathrm{are}\mathrm{the}\mathrm{tangent}\mathrm{lines}\mathrm{at}x=n$$$$$$$$\therefore y_f\mathrm{\perp }{y}_g\because f^{\prime }\left(n\right)\times (-\frac{1}{f^{\prime }\left(n\right)})=-1$$$$\therefore \mathrm{if}g\left(x\right)=\int -\frac{1}{f^{\prime }\left(x\right)}dx,\mathrm{the}\mathrm{tangent}\mathrm{lines}$$$$\mathrm{at}x=n\mathrm{are}\mathrm{perpendicular}$$$$$$

Commented by Filup last updated on 12/Nov/15

$$\mathrm{This}\mathrm{took}\mathrm{me}\mathrm{way}\mathrm{to}\mathrm{long}\mathrm{to}\mathrm{figure}\mathrm{out}$$$$\mathrm{and}\mathrm{prove}\mathrm{haha}.\mathrm{So}\mathrm{simple}$$

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