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April 25th, 2016, 01:08 PM   #1
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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 $\displaystyle \frac{17}2$
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April 25th, 2016, 08:45 PM   #2
jks
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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:

$\displaystyle \left(x^2+\frac{1}{4} \right)x=\frac{17}{2}$

$\displaystyle x=2$
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