Hi all. My problem is to find the equation of the tan line to f(x)=tan(x)+9* cos(x) at x=pi. So I know that the formula for the equation of the tan line is f(a)+f'(a)(x-a). I have tan(pi)+9*cos(pi)= 0+9(-1)=-9. The next step is to find the derivative of f(x) which is f'(x)=sec squared x-(9sin(x)). I don't think I know how to work with the sec squared x. My initial thought was to use 1+tan squared x. Using the 1+tan squared x I could then use the property 1-cos(2x)over 1+cos(2x). I do not know if I am going in the right direction. Please advise.
I am stuck on finding the equation of the Tan line
- Thread starter mikagurl
- Start date
all you have to do is find the first derivative which i think its a little bit complicated the way you started i will show you this one i hope it will help you :
f'(x)=(tan(x))'+(9* cos(x))'=1/(cos squared x)-9 sin x now just calculate for x=pi and you will have :
f'(pi)=(1/(-1)^2)-0=1
so for the tan line implies : y= x-9-pi
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- Feb 4, 2004
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Actually, no. The first derivative will only provide the slope at the point of tangency. In order to find the tangent <em>line</em>, the poster needs to do (almost) exactly as she showed: find the point of tangency, (a, f(a)), and then plug this, along with the slope, f'(a), into the line equation, y = f'(a)(x - a) + f(a).
Almost. The point of tangency will be the point (x, y) on the curve at a = \(\displaystyle \pi.\) So the point will be (x, y) = (a, f(a)) = \(\displaystyle \left(\pi,\, \tan(\pi) \, +\, 9\, \cos(\pi)\right).\) So evaluate this to get the point at which the line will be tangent. Then you'll need the slope at that point, which will be (as you've noted) m = f'(a). Then the line equation will be y - f(a) = m(x - a), or y = f'(a)(x - a) + f(a). (You can't forget the "y=", or else you don't have a line "equation"!)
Yes. This means that the point of tangency (where the straight line is going to touch) is at (x, y) = (a, f(a)) = \(\displaystyle (\pi,\, -9).\)
So don't! You know that the secant is the reciprocal of the cosine, so flip, evaluate the cosine, square, and simplify.
. . . . .\(\displaystyle \sec^2(x)\, -\, 9\, \sin(x)\, =\, \dfrac{1}{\left(\cos(x)\right)^2}\, -\, 9\, \sin(x)\)
Plug in \(\displaystyle \pi\), evaluate, and simplify. This gives you the value of f'(a) = m, the slope of the tangent line. Plug this into the line formula, and you're done (other than simplifying).