Menu Close

Lets-say-we-have-three-points-A-0-0-B-x-y-C-x-y-Assuming-that-both-B-and-C-are-point-on-a-fuction-y-f-x-we-can-calculate-the-area-under-the-point-where-it-makes-a-right-triangle-with-the-o




Question Number 4535 by FilupSmith last updated on 05/Feb/16
Lets say we have three points:  A(0, 0)  B(x, y)  C(δx, δy)    Assuming that both B and C are point  on a fuction y=f(x), we can calculate  the area under the point where it makes  a right triangle with the origin and x−axis.    Can we calculate the area of ABC?
$$\mathrm{Lets}\:\mathrm{say}\:\mathrm{we}\:\mathrm{have}\:\mathrm{three}\:\mathrm{points}: \\ $$$${A}\left(\mathrm{0},\:\mathrm{0}\right) \\ $$$${B}\left({x},\:{y}\right) \\ $$$${C}\left(\delta{x},\:\delta{y}\right) \\ $$$$ \\ $$$$\mathrm{Assuming}\:\mathrm{that}\:\mathrm{both}\:{B}\:\mathrm{and}\:{C}\:\mathrm{are}\:\mathrm{point} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{fuction}\:{y}={f}\left({x}\right),\:\mathrm{we}\:\mathrm{can}\:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{under}\:\mathrm{the}\:\mathrm{point}\:\mathrm{where}\:\mathrm{it}\:\mathrm{makes} \\ $$$$\mathrm{a}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{and}\:\mathrm{x}−\mathrm{axis}. \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:{ABC}? \\ $$
Commented by FilupSmith last updated on 05/Feb/16
Commented by Yozzii last updated on 07/Feb/16
tanα=y/x, tanφ=δy/δx  ⇒α−φ=tan^(−1) (y/x)−tan^(−1) (δy/δx)  A_(△ABC) =(1/2)(√((y^2 +x^2 )({δy}^2 +{δx}^2 )))[sin{tan^(−1) (y/x)−tan^(−1) ((δy)/(δx))}]
$${tan}\alpha={y}/{x},\:{tan}\phi=\delta{y}/\delta{x} \\ $$$$\Rightarrow\alpha−\phi={tan}^{−\mathrm{1}} \left({y}/{x}\right)−{tan}^{−\mathrm{1}} \left(\delta{y}/\delta{x}\right) \\ $$$${A}_{\bigtriangleup{ABC}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({y}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)\left(\left\{\delta{y}\right\}^{\mathrm{2}} +\left\{\delta{x}\right\}^{\mathrm{2}} \right)}\left[{sin}\left\{{tan}^{−\mathrm{1}} \frac{{y}}{{x}}−{tan}^{−\mathrm{1}} \frac{\delta{y}}{\delta{x}}\right\}\right] \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *