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Calculus/Calc. HELP!!!!

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Expert: - 2/4/2010


Question
1.given y=x^2-2x+3 find the equation for the line which contains (2,3) and is perpendicular to the tangent of the parabola (2,3)

2.find the area of that part of the 1st quadrant which lies below both the line in #1 and the parabola

Answer
1. If the slope of a line is m, the perpendicular slope is -1/m.
The slope of a tangent has the derivative at a slope, and y' = 2x- 2.

Since this is at the point x = 2, the slope is m = -1/y'(2) = -1/(4-2) = -1/2.

The point slope form of a line is y-y0 = m(x-x0).

In this case, m = -1/2, x0=2, and 0=3.


2. If we set y to 0 in the line it will tell us where it crosses the axis.
It will be at x0 - y0/m.

Compute the distance from 2 to this point, multiply by 3, and divide by 2.
That is the area of the triangle.

Evaluate ∫x² - 2x + 3 dx from 0 to 2 to find the area on the other side.

Add them together to find the total area.

Note that the integral is F(x) = x³/3 - x² + 3x, and the result desired is F(2) - F(0).

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