In Problems 92-98, use Descartes's method from Problem 91 to find the equation of the line tange

In Problems 92-98, use Descartes's method from Problem 91 to...

Key Concepts

Descartes's Method

Descartes's method refers to a geometric and algebraic approach attributed to René Descartes for determining tangents to curves. In the context of calculus, this method incorporates concepts of differentiation and algebraic manipulation to derive the equation of a tangent line, thereby linking historical approaches with modern techniques in calculus.

Slope as the Derivative

The derivative of a function at a specific point gives the slope of the tangent line at that point. This concept is vital because once the derivative is computed—often using implicit differentiation for curves defined by general equations—it allows one to formulate the equation of the tangent line using the point-slope form.

Equation of a Line

The equation of a line in point-slope form (y - y? = m(x - x?)) is a fundamental concept used to express the tangent line once the slope at the point of tangency is known. This formulation provides a direct way to describe the line that best approximates the curve at a particular point.

Tangent Line

A tangent line to a curve at a given point is a straight line that just touches the curve at that point and has the same instantaneous rate of change as the curve. The tangent line is significant because it provides a linear approximation of the curve near the point of tangency and is used in many areas of calculus and geometric analysis.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of functions that are defined by equations in which the dependent variable is not isolated on one side. By differentiating both sides of the equation with respect to the independent variable and applying the chain rule where necessary, one can solve for the derivative of the dependent variable, which is critical for analyzing rates of change and slopes of curves.

Transcript

00:01 In this problem, we're asked to write the equation of the tangent line that goes through the point one three from this equation of the circle.

00:10 So using the decartis method, we know that the system of the equation x squared plus y squared equals 10.

00:20 And then the second equation is the equation of the tangent line, which is in the form of y equals to mx plus b.

00:28 So we need to find the slope and the y intercept.

00:31 So in order to find slope and y intercept, we need to rewrite the b in terms of slope so that when we find the slope, then we can plug in to find what the y intercept is.

00:43 So we can use this point one -three plug in here, where 3 is the y, and m times 1 is still m plus b.

00:51 So therefore, your y intercept is 3 minus m.

00:54 So therefore, my system of the equation is going to be, y equals to mx plus b, which is 3 minus m.

01:04 So now i'm going to combine these two equations, and we know what y equals to, so we're going to substitute this into the previous equation there.

01:15 So the first equation is going to become x squared plus mx plus 3 minus m squared squared equals to 10.

01:31 So now we're going to expand this and i'm going to leave mx as together and going to leave 3 minus m together.

01:42 So remember that if you have a squared plus b square, sorry, if you have a plus b squared, that's the same thing as a squared plus 2a b plus b squared.

01:57 So i'm going to leave a as the mx and then b as 3.

02:02 Minus m and we can expand it towards the end.

02:06 So therefore this equation is going to give us x squared plus m squared x squared plus there's going to be two of times a times b so that's going to be give you 3m times 3 minus m x and then plus 3 minus m squared and then i'm going to go ahead move the 10 to the other side, so minus 10, and it equals to 0.

02:43 So now what i have is, you can see that this is kind of like a quadratic equation.

02:48 If i combine, if i factor out the x squared, that's going to give me 1 plus m squared, plus 2m times 3 minus m x, and then plus the constant, which is going to be 3 minus m squared, minus 10 equals to 0.

03:14 So the number beside x square is a, and the number in front of x is b, and then this is your constant.

03:23 Okay, so sorry, i put the square in the wrong place, square out here.

03:28 So this part here is the constant.

03:32 So using the dicartis method, we know that in order to find the slope, we're going to use the discriminant, which is b squared minus 4 ac and must equal to 0 because there can only be one solution for the slope.

03:49 So we're going to use this formula b squared minus 4 ac to solve for m.

03:53 So my b here is 2m times 3 minus m squared minus 4 times a.

04:05 A is 1 plus m squared times c.

04:13 C is that whole thing, which is 3 minus m squared minus 10, and that's equal to zero.

04:31 So we need to simplify this.

04:33 And so let's simplify what's inside that parentheses, three minus, i keep writing the square in the wrong place.

04:41 It should be square here.

04:42 Okay.

04:43 So we need to distribute the 3 minus m squared and then subtract 10, and then we have to multiply by 1.

04:50 Plus m squared and then distribute by four.

04:53 So there are three steps here.

04:55 So the first thing i'm going to do is foil three minus m square out here.

05:00 So three minus m square gives you nine minus six m plus m squared.

05:07 And then you have to subtract 10.

05:10 So that's going to give you m squared minus six m.

05:14 Nine minus ten is going to give you negative one.

05:17 So therefore now we have here.

05:19 So i'm going to go ahead and square the 2m, so that's going to be 4m squared times 3 minus m squared minus 4 times 1 plus m squared times we have m squared minus 6m minus 1 equals to 0.

05:44 So now we're going to simplify this even more.

05:47 So we have 4m squared...

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