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    havj0y's Avatar
    havj0y Posts: 7, Reputation: 1
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    #1

    Oct 20, 2011, 01:05 AM
    Find the equation normal to y=e^(-x^2)
    at the point where x=1
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Oct 20, 2011, 06:33 AM
    Method:

    1. Find the value for y at x = 1. Let's call this point (x_1, y_1)
    2. Find the slope of y(x) at x = 1.
    3. The slope of the line that is perpendicular to y(x) at x = 1 has a slope that is the negative inverse of the slope of y(x) at that point. Hence m (slope of the line) is equal to negative 1 divided by the answer you found in step 2.
    4. Now that you know the slope of the perpendicular line and the fact that it goes through the point (x1,y1), the formula for the line is:



    You can rearrange this to get it into standard form for a line: y = mx + b.

    Hope this helps. Post back with the answer you get and we'll check it for you.
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