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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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