(a) Find all points on y^2+x y+x^2=1 with x=1 (b) Find d y /...

Key Concepts

Tangent Line Slope

The slope of the tangent line to a curve at a specific point is given by the derivative of the function evaluated at that point. In the context of implicit differentiation, after finding an expression for dy/dx, the slope of the tangent line at points of interest is obtained by substituting the coordinates into this derivative expression.

Implicit Differentiation

This is a method used for finding the derivative of a function when the function is not given explicitly as y = f(x) but rather defined implicitly by an equation involving both x and y. Implicit differentiation involves differentiating both sides of the equation with respect to x, treating y as a function of x, and applying the chain rule where necessary to obtain dy/dx.

Substitution Method

This concept involves plugging in a given value for one of the variables into the equation defining the curve. It is used to find the corresponding values of the other variable(s) that satisfy the equation. In this problem, substituting x = 1 leads to solving for y and finding specific points on the curve.

Solving Quadratic Equations

When substitution results in a quadratic equation in one variable, the techniques for solving quadratic equations become essential. These techniques, such as factoring, completing the square, or using the quadratic formula, help in determining the possible values of the variable, and thereby, finding the intersection points or specific values on the curve.

Transcript

0:00 All right.

00:01 We are just doing a bunch of stuff with the implicitly defined function.

00:06 Y squared plus xy plus x squared is equal to 1.

00:13 The first part for part a is to figure out the ordered pair when x equals 1.

00:20 Well, i can use direct substitution with this problem and replace, i can tell.

00:29 All i'm doing is taking this x is 1.

00:31 I'm going to substitute in here and here and figure out where y could possibly equal.

00:39 So if i subtract the 1 over, well, both of those will cancel and i'm left with 0 on the right side.

00:44 I can factor out a y of the remaining pieces and be left with y plus 1.

00:50 So we have two possibilities.

00:53 We have the ordered pair that y could be 0 or y could equal negative 1.

00:57 So the ordered pairs that we're considering is 1 ,0, and 1.

01:03 Negative 1 because the x values the same throughout so there's part a part b is asking you to find d y d y d x well that's the implicit part the differentiation where you do the derivative of y squared is 2 y d y d y d x notice that this middle piece is actually the product rule or the derivative of x would be one you leave the y alone plus now you leave the x alone the derivative of y would be d -y -d -x.

01:34 The derivative of x squared is 2x, and then the derivative of 1, a constant, is equal to 0.

01:42 So i would actually do two steps at once, and hopefully you can follow, that you can factor out a d -y -d -x for these two terms...

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