# Implicit differentiation.

• Nov 10th 2009, 07:22 AM
Rumor
Implicit differentiation.
There's a multi-step problem that I need help in figuring out how to do.

"a) Find the equation of the tangent lines to the circle $x^2+y^2=25$ at the points where $x=4$.

b) Find the equations of the normal lines to this circle at the same points. (The normal line is perpendicular to the tangent line at this point.)

c) At what point do the two normal lines intersect?"

Could someone tell me how to start the solution to these questions, please; i.e., what do do? Thank you.
• Nov 10th 2009, 07:40 AM
tom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/asy/i...iff/circle.png

... where

http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (which is the inner function of the composite, which needs the chain rule).

This is called implicit differentiation, because you use the chain rule to differentiate y with respect to x. Solve the bottom row for y' i.e. dy/dx, and then plug in 4 for x and whatever values you are able to determine for y when x is 4, to get the two values of dy/dx at those points. That's the slopes of the tangents at those points, so multiply them by (what number?) to get the slopes perpendicular to those. Then you at least have the slopes of the normals. Then you can review the situation and see if you have the information to deduce the equations of these lines.

__________________________________________
Don't integrate - balloontegrate!

http://www.ballooncalculus.org/forum/top.php Calculus Forum

http://www.ballooncalculus.org/asy/doc.html Calculus Drawing with LaTeX and Asymptote!
• Nov 10th 2009, 07:42 AM
Quote:

Originally Posted by Rumor
There's a multi-step problem that I need help in figuring out how to do.

"a) Find the equation of the tangent lines to the circle $x^2+y^2=25$ at the points where $x=4$.

b) Find the equations of the normal lines to this circle at the same points. (The normal line is perpendicular to the tangent line at this point.)

c) At what point do the two normal lines intersect?"

Could someone tell me how to start the solution to these questions, please; i.e., what do do? Thank you.

The equation of a tangent line to a curve has the form $\Delta y=y'\Delta x$, so you need to find $y'$, and you need to find the points. The points are easy to find, just plug in $x=4$ in the equation $y=\pm\sqrt{25-x^2}$.

Implicity differentiate the equation:
$x^2+y^2=25$

Differentiating gives:

$2x+2yy'=0$

$y'=-\frac{x}{y}$

Do you know how to write the equation of the tangent lines with this information? Since $y=\pm\sqrt{25-x^2}$, there are two values of $y$ for every $x$. So $x=4$ will give two points. Since you also have the equation $y'=-\frac{x}{y}$, you will know the slope of the lines through those points. It would be a good idea to draw the circle on the xy-plane to visualize this.

Let me know if I lost you anywhere.
• Nov 10th 2009, 08:09 AM
Rumor
Thanks to the both of you!

That helped me figure out the tangent lines, but I'm still confused on how to find the normal lines from this.
• Nov 10th 2009, 08:24 AM
tom@ballooncalculus
Well, if y' at (4,3) is -x/y = -4/3, then the slope of the normal at this point is perpendicular to that, i.e. minus the reciprocal of that... and like the tangent at (4,3) it passes through (4,3), so you can deduce the rest of the equation using 'point-slope' or whatever your favourite method... ...? (e.g. just reasoning...)