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 April 25th, 2016, 02:08 PM #1 Newbie   Joined: Apr 2015 From: Slovakia Posts: 5 Thanks: 0 Finding point in graph Hey! Anyone has any idea how to solve this problem ? Find point on graph of function $\displaystyle f(x)=x^2-1$ so that surface area of triangle which is made of axis $\displaystyle O_y$ and tangent and normal, is in wanted point equal to $\displaystyle \frac{17}2$
 April 25th, 2016, 09:45 PM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 633 Thanks: 94 Math Focus: Electrical Engineering Applications Hi Mrto, I think that this is right: Confining ourselves to positive values of x, at a point (x,y), picture the triangle formed by the y axis, the tangent to the curve, and the normal to the curve. The value of y really does not need to be calculated. Notice that since the derivative is 2x and x is positive, the tangent will intersect the y axis below the value of y by a (positive) distance of $\Delta y_l$. The normal will have a slope of $-\frac{1}{2x}$ and the normal will intersect the y axis a (positive) distance of $\Delta y_h$ above y. Since $\text{slope}=\frac{\Delta y}{\Delta x}, \quad \Delta y=\text{slope} \cdot \Delta x$ and given the definitions $\Delta x = x$ (positive) so: $\displaystyle \Delta y_l=2x \cdot x=2x^2$ $\displaystyle \Delta y_h=\frac{1}{2x} \cdot x=\frac{1}{2}$ The base of the triangle is: $\displaystyle \Delta y_l + \Delta y_h=2x^2+\frac{1}{2}$ The height of the triangle is $x$ and the area is: $\displaystyle \frac{1}{2}\cdot \text{base} \cdot \text{height}=\frac{1}{2} \left(2x^2+\frac{1}{2} \right)x=\left(x^2+\frac{1}{4} \right)x$ so: $\displaystyle \left(x^2+\frac{1}{4} \right)x=\frac{17}{2}$ $\displaystyle x=2$ Thanks from Mrto

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