(a) find two explicit functions by solving the equation for y in terms of x,(𝐛) sketch the graph

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

Key Concepts

Graphical Analysis of Conic Sections

Graphical analysis of conic sections involves studying the curves defined by second-degree equations, such as circles, ellipses, parabolas, and hyperbolas. Understanding the geometric features of these curves—like their centers, radii, axes, and vertices—helps in accurately sketching the graphs and analyzing their behavior. This concept is critical for interpreting and visualizing the solutions of quadratic equations.

Chain Rule in Differentiation

The chain rule is a fundamental principle in calculus used to differentiate composite functions. When differentiating a function defined implicitly, the chain rule allows one to correctly account for the dependency of the inner function on the independent variable. It ensures that the derivative of the overall function is computed accurately, especially when multiple functions are composed.

Implicit Differentiation

Implicit differentiation is a method used to find the derivative of a function defined by an implicit relationship between variables. Instead of isolating the dependent variable, the derivative is obtained by differentiating both sides of the equation with respect to the independent variable and then solving for the derivative. This technique is important when dealing with equations that are difficult or impossible to solve explicitly.

Completing the Square

This algebraic technique rewrites quadratic expressions into a perfect square plus a constant. It is particularly useful for transforming conic section equations into their standard forms, which makes it easier to identify properties such as the center, vertex, or radius of the figure. Completing the square is essential when converting implicit equations into forms that reveal the structure of the graph.

Implicit to Explicit Conversion

This process involves solving an equation in which the dependent variable is intermixed with the independent variable, to express the dependent variable explicitly as a function of the independent variable. Having explicit functions facilitates analysis, such as graphing and differentiation, as each branch of the curve can be studied separately.

Transcript

00:01 For this problem, we start with the equation x squared plus y squared minus 4x plus 6y plus 9 equals 0.

00:06 And our first step is, or the first part of the problem, is that we want to find two explicit functions for y in terms of x.

00:12 So the first thing that i'm going to do here is rearrange the equation.

00:16 So i get all the y terms on one side, or all the y terms and are constant on one side, and the x terms on the other.

00:23 So we'll have y squared plus 6y plus 9 equals negative x squared plus 4x.

00:31 Then i'm going to try to complete the square on the left -hand side.

00:35 So we'd have that this is going to be y plus 6 over 2 times 1, so y plus 3, all squared, plus c, so plus 9, minus b squared, 36 over 4 times a.

00:48 So 36 over 4 is going to give us 9.

00:53 So we'd have plus 9 minus 9, giving us just y plus 3 all squared.

00:58 And then on the right -hand side, we'd have that this is going to to equal negative x, well, we can just leave this as negative x squared plus 4x for now.

01:07 Then we can take the square root of both sides, in which case we get y plus 3 equals plus or minus the square root of negative x squared plus 4x.

01:19 And then we need to subtract 3 from both sides, in which case we get y equals plus or minus the square root of negative x squared plus 4x minus 3.

01:32 To aid in due, our graphing, i'll also note that this can be put into a different form if we complete the square on the right -hand side.

01:41 So we would have negative of x, then it would be plus 4 over 2 times negative 1, so that would be minus 2 all squared, plus c, so plus 0, minus b squared, so minus 16 over 4a, so that would be plus 16 over 4, or plus 4, which then means that we can rearrange the overall equation into being y plus 3 all squared plus x minus 2 all squared equals 4 so that for part b we are asked to sketch the graph so we can note that we have from that alternate form that's clearly the equation of a circle with radius 2 and a center at y equals negative 3 and x equals positive 2 so we'd have that the positive root that we get here is going to correspond to the positive solution for y or expression for y and the negative or the lower half of the circle is from the negative solution for y then for part c we're asked to differentiate the explicit functions so we'll have d y by d x it's going to equal well actually i'll write this as d y plus so differentiating the positive root we'd need to apply chain rule so differentiating negative x squared plus 4x with respect to x would give us negative 2x plus 4 times 1 half of negative x squared plus 4x to the power of negative 1 half which we can then write as being negative x plus 2 over the square root of negative x squared plus 4x and the negative root for y we would have essentially the same thing only the signs up top would be reversed, so we'd have x minus 2 over the square root of negative x squared plus 4x.

03:42 Lastly, for part d, we're asked to find the derivatives using explicit or implicit differentiation instead...

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