(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). $$ 4 y^{2}-x^{2}=4 $$

Key Concepts

Implicit Differentiation

Implicit differentiation is used when the relationship between x and y is given in implicit form and cannot be easily solved for y. By differentiating both sides of the equation with respect to x and then solving for dy/dx, one can obtain the derivative without explicitly isolating y. This method often involves applying the chain rule due to the dependence of y on x.

Solving for Explicit Functions

This concept involves transforming an implicit equation into explicit forms by isolating one variable in terms of the other. In many cases, like with equations involving squares, this process introduces multiple branches corresponding to the positive and negative square roots. Understanding this step is crucial in recognizing the complete set of explicit solutions that describe the relationship between the variables.

Graphing Conic Sections

Graphing the equation entails interpreting it as a conic section, which in this context results in a curve (such as a hyperbola, ellipse, or parabola). Identifying the key features, such as vertices, axes, and symmetry from the explicit functions, is fundamental for accurately sketching the graph. This also helps in visually understanding the domains and ranges of the resulting functions.

Explicit Differentiation

Explicit differentiation is the process of differentiating a given function that has been expressed in the form y = f(x). Once the equation has been solved explicitly, standard differentiation techniques can be applied to find the derivative of y with respect to x, treating it as an ordinary function.

Equivalence of Derivatives

This concept stresses that the derivative obtained through explicit differentiation should match the derivative computed via implicit differentiation. Establishing this equivalence serves as a consistency check, affirming that both methods yield the same rate of change and thus correctly describe the behavior of the function.

Transcript

00:01 For this problem, in part a, we are asked to find two explicit functions by solving the equation 4 y squared minus x squared equals 4 for y in terms of x.

00:09 So this is a pretty simple one to rearrange.

00:12 We can isolate y squared by adding x squared to both sides.

00:16 So we have 4 y squared equals 4 plus x squared.

00:19 And then divide both sides by 4, giving us y squared equals 4 plus x squared over 4.

00:25 And then we take the square root of everything, giving us that y equals plus x squared.

00:30 Or minus.

00:32 We can write this as, oh, actually, we'll get ahead of myself here, y equals plus or minus the square root of 4 plus x squared all over 4, which we can then rewrite as y equals plus or minus 1 over 2 times the square root of 4 plus x squared.

00:52 Then for part b, we are asked to sketch the graph of the equation and label the parts given by the corresponding explicit functions.

01:00 So i'm not going to sketch this out, instead, i'm just going to put in an exact plot here.

01:06 So we can see that we have a hyperbola here, where we have the, i believe those are the vertices at plus or minus y equals 1.

01:19 And we can see that the positive will be from the y plus, as i'll label it, and the negative values will be from the y minus.

01:27 For part c, we're asked to differentiate the explicit functions.

01:30 And, oh, actually, that is part c.

01:34 So i'll write this as d .y by dx for...

01:40 First, i'll just write for y plus...

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