Expert: Scotto - 11/8/2008

Question
I have one question about the equation of the normal line:

1)  Let (3+tan(x))/(sin(x)).  Write the equation of the normal line to f(x) at x=pi/4.  Please give exact answers and no decimals.

Thanks for you help,
robert

Answer
The equation to the line must have the same slope where it contacts the curve.

The slope of where it contacts the curve is the derivative of the curve.

The derivative is a quotient rule with f(x)=3+tan(x) and g(x)=sin(x).
f'(x)=sec²(x) and g'(x)=cos(x).

Remembering the quoteint rule is
(low d high - high do low)/low² and that high is f(x) and low is
g(x), the substituions can be made.

The result is (sin(x)sec²(x) - (3+tan(x))cos(x))/sin²(x), which can be simplified.

I find the best way for me to simplify is to convert everything to sin(x) and cos(x).  Note that sec(x) = 1 / cos(x) and that tan(x) = sin(x)/cos(x).

Substituting these back in gives
(sin(x)/cos²(x) - 3cos(x) - (sin(x)/cos(x))cos(x))/(cos(x)).

Dividing every term by the cos(x) at the end gives
sin(x)/cos^3(x) - 3 - sin(x)/cos(x).

Simplifying the last term gives
tan(x)sec²(x) - 3 - tan(x).

Now that you have the derivative, the π/4 can be taken and put into it to find the slope.  Note that π/4 is 45º Note that
tan(π/4) = 1 and sec²(π/4) = √2.

Put the values of these trig function at π/4 back in the last equation { tan(x)sec²(x) - 3 - tan(x) } and the slope can be found.

The equation for a line passing through a point (x0,y0) with a slope m is y-y0 = m(x-x0).

Once this has been found, the m can be multiplied out and the y0 added to both sides, so that the equation of the line is in the proper form y = mx + b.  Note that here b = -m*x0 + y0.