(a) find two explicit functions by solving the equation for...

(a) find two explicit functions by solving the equation for $y$ in terms of $x$, (b) sketch the graph of the equation and label the parts given by the corresponding explicit functions, (c) differentiate the explicit functions, and (d) find $d y / d x$ implicitly and show that the result is equivalent to that of part (c). $$ x^{2}+y^{2}-4 x+6 y+9=0 $$

Key Concepts

Solving for Explicit Functions

In the context of implicit equations, especially those containing quadratic terms, it is often necessary to solve for one variable explicitly in terms of the other. This involves isolating the variable (for example, y) and taking square roots, which usually results in two distinct functions representing different branches of the original curve.

Implicit Differentiation

This is a technique used to differentiate equations that define one variable in terms of another indirectly. Instead of first solving for one of the variables explicitly, differentiation is applied directly to both sides of the equation, using the chain rule as necessary, making it particularly useful when dealing with relations that are difficult to solve explicitly.

Completing the Square

This procedure involves rewriting a quadratic expression as a perfect square plus or minus a constant. It is an essential algebraic manipulation used to convert general quadratic equations—especially those representing conic sections—into their standard forms, which makes it easier to identify key features such as the center, vertex, or radius.

Graphing Conic Sections

Graphing conic sections involves recognizing the standard forms of the equations after algebraic manipulation (such as completing the square) and then using the identified parameters (like center, radius, or vertex) to accurately depict the curve. This visual understanding assists in analyzing the behavior of the function and its derivatives.

Transcript

00:01 For this problem, we are asked in part a to find two explicit functions by solving the equation given for y in terms of x.

00:09 And so for that, let's see here, we'll have to end up effectively first completing this square for this expression.

00:19 Specifically, we'll be doing that, or i'll do that, for the y terms and the x terms separately.

00:25 So we have y squared minus 6y would have to be equal to y minus 3 squared.

00:36 And if that was the case, then we would expect to have a plus 9.

00:39 So we need to have a minus 9 that we put out front.

00:44 Then for x, we can see we have x squared minus 4x.

00:49 So to get that, we'd have to have x minus 2 squared.

00:54 And then if that were the case, we would expect to have a 2 squared.

00:56 Sitting out front, which would be a plus 4, so we'll have to subtract off 4, which means that we can write our expression up above as y minus 3 squared minus 9 plus 9, so that adds up to 0, equals, or excuse me, not equals.

01:15 Well, actually, i won't get ahead of myself.

01:17 So it would be plus x minus 2, squared, minus 4 equals 0.

01:24 So we can then write that this would be y minus 3, squared equals 4 minus x minus 2 x minus 2 squared then we take the square root of both sides we have y minus 3 equals the square root of 4 minus x minus 2 squared which then means we have y equals 3 plus the square root of 4 minus x minus 2 squared.

01:58 Actually, excuse me, not just plus.

02:00 I missed the plus or minus here.

02:02 So 3 plus or minus root 4 minus x minus 2 squared.

02:05 And i'll note that for part b, where we are asked to sketch, it would actually be easier to write this in a more familiar form, specifically y minus 3 squared plus x minus 2 squared equals 4.

02:22 So for part b, we can read off easily that that's the equation of a circle centered at the point 3 -2.

02:30 So that's x equals 2, y equals 3...

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