Calculus Help

[flsweetheart422]

I have no idea how to solve this problem. I have the final answer, but don't know how to get there. I think I need to use implicit differentiation, but I'm not sure, and I'm quite confused.

Any help would be greatly appreciated.

Find the equation of the tangent line to the curve is x^2+y^2 = 169 at the point (5, -12).

A. 5y-12x=-120

B. 5x-12y=119

C. 5x-12y=169

D. 12x+5y=0

E. 12x+5y=169

I have been told that the final answer is (B)5x-12y=119, but none of my friends seem to be able to explain it.

Edited November 30, 2008 by flsweetheart422

[Abu]

2x + 2y(dy/dx) = 0

(dy/dx) = -2x/2y = -x/y

(Just to confirm I'm not rusty: http://www.mathway.com/answer.aspx?p=ccal?...y/dx?p=0?p=?p=)

Gradient of tangent line = 5/12

Edit: http://www.webmath.com/cgi-bin/equline.cgi...k=equline2.html

Therefore: y=5/12x-169/12

Multiply through by 12 and you should get your answer.

[SharkSpider]

Well, you've got a circle with radius 13 that has no translation from the origin.

The key to implicit differentiation is that you take d/dx of every term. Then, you need to use the chain rule on all terms that contain y, so that d/dx of y^2 is 2y times dy/dx.

Remember that with equations of y = something, differentiating gives you the slope of the tangent line, and that really, all you're doing is implicitly differentiating, because d/dx of y is 1 times dy/dx, which is what you get when you differentiate normal functions.

So, differentiating...

x^2 + y^2 = 169 gives:

2x + 2y(dy/dx) = 0, so dy/dx = -x/y.

Or, 5/12.

y - b = m(x - a) For point-intercept format.

So:

y - (-12) = 5/12(x - 5)

y = 5x/12 - 25/12 - 12

12y = 5x - 25 - 144

12y - 5x = -169

5x-12y=169

So the answer they gave you was wrong. The answer should be C.

[flsweetheart422]

Thanks guys! I appreciate it!

[rofler]

You don't really need calculus at all for this question. Any tangent line to a circle will be perpendicular to the line through the point of tangency and the center of the circle. This line passes through the center $(0,0)$ and the point of tangency, $(5,-12)$ so will have equation $y=-(12/5)x$.

Then, the line of tangency is perpendicular to this line, so has slope $5/12$, and passes through $(5,-12)$ so has equation $y=(5/12)(x-5)-12$ or $5x-12y=169$.