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

With linear functions f(x) and g(x),  if f(x)⊥g(x), then:  m_f m_g =−1    where m_i  is the gradient  of function i(x).    Does that therefore mean that, if given  function (including non−linear) f(x),  f′(x)g′(x)=−1,     if f(x)⊥g(x) at x=n?    e.g.  f(x)=x^2   f′(x)=2x    if f(x)⊥g(x) at x=n:  ∴f′(x)g′(x)=−1  g′(x)=−(1/(2x))  g(x)=−∫(1/(2x))dx  ∴g(x)=−(1/2)ln(2x)+c_1     f(x)=x^2 ,   g(x)=−(1/2)ln(2x)  ∴ x^2 (−(1/2)ln(2x))=−1    My question therefore comes down to:  Are the tangent lines of f(x) and g(x)  at x=n perpendicular to each other  for all values of n∈{f(x), g(x)}?

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

Answered by Filup last updated on 12/Nov/15

eqn of tangent line on f(x) at x=n:  y_f =f′(n)(x−n)+f(n)  ∴egn of line on g(x) at x=n:  f′(x)g′(x)=1  y_g =g′(n)(x−n)+g(n)  ∴y_g =−(1/(f′(x)))(x−n)+g(n)    y_f =f′(n)(x−n)+f(n)  y_g =−(1/(f′(n)))(x−n)+g(n)    y_f  and y_g  are the tangent lines at x=n    ∴y_f ⊥y_g ∵f′(n)×(−(1/(f′(n))))=−1  ∴if g(x)=∫−(1/(f′(x)))dx, the tangent lines  at x=n are perpendicular

$${eqn}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{on}\:{f}\left({x}\right)\:\mathrm{at}\:{x}={n}: \\ $$$${y}_{{f}} ={f}'\left({n}\right)\left({x}−{n}\right)+{f}\left({n}\right) \\ $$$$\therefore{egn}\:\mathrm{of}\:\mathrm{line}\:\mathrm{on}\:{g}\left({x}\right)\:\mathrm{at}\:{x}={n}: \\ $$$${f}'\left({x}\right){g}'\left({x}\right)=\mathrm{1} \\ $$$${y}_{{g}} ={g}'\left({n}\right)\left({x}−{n}\right)+{g}\left({n}\right) \\ $$$$\therefore{y}_{{g}} =−\frac{\mathrm{1}}{{f}'\left({x}\right)}\left({x}−{n}\right)+{g}\left({n}\right) \\ $$$$ \\ $$$${y}_{{f}} ={f}'\left({n}\right)\left({x}−{n}\right)+{f}\left({n}\right) \\ $$$${y}_{{g}} =−\frac{\mathrm{1}}{{f}'\left({n}\right)}\left({x}−{n}\right)+{g}\left({n}\right) \\ $$$$ \\ $$$${y}_{{f}} \:\mathrm{and}\:{y}_{{g}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{lines}\:\mathrm{at}\:{x}={n} \\ $$$$ \\ $$$$\therefore{y}_{{f}} \bot{y}_{{g}} \because{f}'\left({n}\right)×\left(−\frac{\mathrm{1}}{{f}'\left({n}\right)}\right)=−\mathrm{1} \\ $$$$\therefore\mathrm{if}\:{g}\left({x}\right)=\int−\frac{\mathrm{1}}{{f}'\left({x}\right)}{dx},\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{lines} \\ $$$$\mathrm{at}\:{x}={n}\:\mathrm{are}\:\mathrm{perpendicular} \\ $$$$ \\ $$

Commented by Filup last updated on 12/Nov/15

This took me way to long to figure out  and prove haha. So simple

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

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