Calculus/Calculus

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Expert: - 9/12/2011


Question
Given y = 1/(X-1) at x=2 find:

a) slope
b) equation of the tangent
c) an equation of the normal
d) draw a graph of the curve, tangent line & normal line in the same square viewing window

Answer
Here, y = (x-1)^(-1).
Since when y = x^n, y' = nx^(n-1), in this case, y' = -1(x-1)^(-2) = -1/(x-1)^2.
Since we have y' = -1/(x-1)^2, put in x=2 and get y' = -1/(2-1)^2 = -1/1^2 = -1/1 = -1.
That is the answer to (a).

The value of y at x=2 is y = 1/(2-1) = 1/1 = 1.
This means the point on the line is (2,1), and we already found the slope to be 1.
The equation is then the point-slope form of a line.
That is y - y0 = m(x-x0) where (x0,y0) = (2,1) and m=1.
Put in the values; multiply by m and drop the parenthesize; add y0 to both sides; that's the equation for (b).

Use the same point, (2,1), but change the slope to -1/m; that answers (c).

To draw a graph, take the x-y graph paper; put a point at (2,1);
draw a line with slope 1 that passes through the point just marked.

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