Implicit Differentiation Tangent Line

5 Key excerpts on "Implicit Differentiation Tangent Line"

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  The How and Why of One Variable Calculus

  • Amol Sasane(Author)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  The process, based on the Chain Rule, is called Implicit Differentiation , and the best way to see this is by looking at an example. Example 4.15 (Tangents and normals to the circle using implicit differentiation) . Let F ( x , y ) : = x 2 + y 2 − 1. Then the point ( x , y ) on the curve described implicitly by F satisfies x 2 + y 2 − 1 = 0. Bearing in mind that y is a function of x , and viewing both sides of the above equation as functions of x , we obtain by differentiating with respect to x that (suppressing the argument of y — that is, writing y instead of y ( x ) everywhere below) 2 x + 2 y · dy dx − 0 = 0, so that if y ( x ) = 0, then dy dx = − x y . So at a point ( x 0 , y 0 ) on the circle, the equation of the tangent line is y − y 0 x − x 0 = − x 0 y 0 , and using x 2 0 + y 2 0 = 1, this can be simplified to yy 0 + xx 0 = 1. Similarly, the equation of the normal line is (assuming x 0 = 0) y − y 0 x − x 0 = y 0 x 0 , that is, y = y 0 x 0 x . ♦ DIFFERENTIATION 153 x 0 y = y 0 x ( x 0 , y 0 ) slope = − x 0 /y 0 F ( x,y ) := x 2 + y 2 − 1 = 0 Example 4.16. Let F ( x , y ) : = y 5 + x 5 − y − 3 x + 2. Then F ( 1, 1 ) = 0. What is the equation of the tangent line to the curve implicitly defined by F at the point ( 1, 1 ) ? We have 5 y 4 dy dx + 5 x 4 − dy dx − 3 = 0 and so dy dx = 3 − 5 x 4 5 y 4 − 1 . So the slope of the tangent line at ( 1, 1 ) is dy dx ( x , y )=( 1,1 ) = 3 − 5 x 4 5 y 4 − 1 ( x , y )=( 1,1 ) = 3 − 5 5 − 1 = − 1 2 . Hence the equation of the tangent line is y − 1 x − 1 = − 1 2 , that is, x + 2 y = 3. ♦ 2 1 0 − 1 − 2 − 1 1 2 x y Exercise 4.33. Find the points on the curve given implicitly by x 2 + xy + y 2 = 9 at which (1) the tangent is parallel to the x -axis (2) the tangent is parallel to the y -axis. 154 THE HOW AND WHY OF ONE VARIABLE CALCULUS Exercise 4.34. Find the tangents to the implicitly defined curve x sin( xy − y 2 ) = x 2 − 1 at the point ( 1, 1 ) .

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  Calculus

  Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  This method of obtaining derivatives is called implicit differentiation. Example 2 Use implicit differentiation to find dy / dx if 5y 2 + sin y = x 2 . d dx [5y 2 + sin y] = d dx [x 2 ] 5 d dx [y 2 ] + d dx [sin y] = 2x 5  2y dy dx  + (cos y) dy dx = 2x The chain rule was used here because y is a function of x. 10y dy dx + (cos y) dy dx = 2x Solving for dy / dx, we obtain dy dx = 2x 10y + cos y (8) Note that this formula involves both x and y. Example 3 Use implicit differentiation to find d 2 y / dx 2 if 4x 2 − 2y 2 = 9. Solution Differentiating both sides of 4x 2 − 2y 2 = 9 with respect to x yields 8x − 4y dy dx = 0 from which we obtain dy dx = 2x y (9) Differentiating both sides of (9) yields d 2 y dx 2 = (y)(2) − (2x)(dy / dx) y 2 (10) Substituting (9) into (10) and simplifying using the original equation, we obtain d 2 y dx 2 = 2y − 2x(2x / y) y 2 = 2y 2 − 4x 2 y 3 = − 9 y 3 In Examples 2 and 3, the resulting formulas for dy / dx involved both x and y. This is not an impediment to finding slopes and equations of tangent lines provided the x- and y-coordinates of the point of tangency are known. 2.7 Implicit Differentiation 105 Example 4 Find the slopes of the tangent lines to the curve y 2 − x + 1 = 0 at the points (2, −1) and (2, 1). Solution Differentiating implicitly yields d dx [y 2 − x + 1] = d dx [0] d dx [y 2 ] − d dx [x] + d dx [1] = d dx [0] 2y dy dx − 1 = 0 dy dx = 1 2y At (2, −1) we have y = −1, and at (2, 1) we have y = 1, so the slopes of the tangent lines to the curve at those points are (Figure 2.7.4) dy dx    x =2 y =-1 = − 1 2 and dy dx    x =2 y =1 = 1 2 2 x y (2, 1) (2, –1) FIGURE 2.7.4 Formula (11) cannot be eval- uated at (0, 0) and hence pro- vides no information about the nature of the Folium of Descartes at the origin.

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  Calculus

  Late Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  59 2 One of the crowning achievements of calculus is its ability to capture continuous motion mathematically, allowing that motion to be analyzed instant by instant. THE DERIVATIVE Many real-world phenomena involve changing quantities—the speed of a rocket, the inflation of currency, the number of bacteria in a culture, the shock intensity of an earthquake, the voltage of an electrical signal, and so forth. In this chapter we will develop the concept of a “derivative,” which is the mathematical tool for studying the rate at which one quantity changes relative to another. The study of rates of change is closely related to the geometric concept of a tangent line to a curve, so we will also be discussing the general definition of a tangent line and methods for finding its slope and equation. 2.1 TANGENT LINES AND RATES OF CHANGE In this section we will discuss three ideas: tangent lines to curves, the velocity of an object moving along a line, and the rate at which one variable changes relative to another. Our goal is to show how these seemingly unrelated ideas are, in actuality, closely linked. TANGENT LINES In Example 1 of Section 1.1 we used an informal argument to find the equation of a tangent line to a curve. However, at that stage in the text we did not have a precise definition of a tangent line. Now that limits have been defined precisely we can give a mathematical definition of the tangent line to a curve y = f (x) at a point P(x 0 , f (x 0 )) on the curve. As illustrated in Figure 2.1.1, the slope m PQ of the secant line through P and a second point Q(x, f (x)) on the graph of f is m PQ = f (x) − f (x 0 ) x − x 0 If we let x approach x 0 , then the point Q will move along the curve and approach the point P. Suppose the slope m PQ of the secant line through P and Q approaches a limit as x → x 0 . In that case we can take the value of the limit to be the slope m tan of the tangent line at P. Thus, we make the following definition. Figure 2.1.1

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  Calculus

  Early Transcendental Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  79 2 One of the crowning achievements of calculus is its ability to capture continuous motion mathematically, allowing that motion to be analyzed instant by instant. THE DERIVATIVE Many real-world phenomena involve changing quantities—the speed of a rocket, the inflation of currency, the number of bacteria in a culture, the shock intensity of an earthquake, the voltage of an electrical signal, and so forth. In this chapter we will develop the concept of a “derivative,” which is the mathematical tool for studying the rate at which one quantity changes relative to another. The study of rates of change is closely related to the geometric concept of a tangent line to a curve, so we will also be discussing the general definition of a tangent line and methods for finding its slope and equation. 2.1 TANGENT LINES AND RATES OF CHANGE In this section we will discuss three ideas: tangent lines to curves, the velocity of an object moving along a line, and the rate at which one variable changes relative to another. Our goal is to show how these seemingly unrelated ideas are, in actuality, closely linked. TANGENT LINES In Example 1 of Section 1.1 we used an informal argument to find the equation of a tangent line to a curve. However, at that stage in the text we did not have a precise definition of a tangent line. Now that limits have been defined precisely we can give a mathematical definition of the tangent line to a curve y = f (x) at a point P(x 0 , f (x 0 )) on the curve. As illustrated in Figure 2.1.1, the slope m PQ of the secant line through P and a second point Q(x, f (x)) on the graph of f is m PQ = f (x) − f (x 0 ) x − x 0 If we let x approach x 0 , then the point Q will move along the curve and approach the point P. Suppose the slope m PQ of the secant line through P and Q approaches a limit as x → x 0 . In that case we can take the value of the limit to be the slope m tan of the tangent line at P. Thus, we make the following definition. Figure 2.1.1

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  Calculus Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  Solution Since the derivative of y with respect to x is y ′ (x) = d dx [ 5x −1 − 1 5 x ] = d dx [5x −1 ] − d dx [ 1 5 x ] = −5x −2 − 1 5 the slope of the tangent line at the point (5, 0) is y ′ (5) = − 2 5 . Thus, the equation of the tangent line at this point is y = − 2 5 (x − 5) Since the y-intercept of this line is 2, the right triangle formed from the coordinate axes and the tangent line has legs of length 5 and 2, so its area is 1 2 (5)(2) = 5 (Figure 2.3.4). 1 2 3 4 5 6 1 2 3 4 5 x y y = 5x –1 – x 5 1 ▴ Figure 2.3.4 Higher Derivatives The derivative f ′ of a function f is itself a function and hence may have a derivative of its own. If f ′ is differentiable, then its derivative is denoted by f ′′ and is called the second derivative of f. As long as we have differentiability, we can continue the process of differentiating to obtain third, fourth, fifth, and even higher derivatives of f. These successive derivatives are denoted by f ′ , f ′′ = ( f ′ ) ′ , f ′′′ = ( f ′′ ) ′ , f (4) = ( f ′′′ ) ′ , f (5) = ( f (4) ) ′ , … If y = f (x), then successive derivatives can also be denoted by y ′ , y ′′ , y ′′′ , y (4) , y (5) , … 116 Chapter 2 / The Derivative Other common notations are y ′ = dy dx = d dx [ f(x)] y ′′ = d 2 y dx 2 = d dx [ d dx [ f(x)] ] = d 2 dx 2 [ f(x)] y ′′′ = d 3 y dx 3 = d dx [ d 2 dx 2 [ f(x)] ] = d 3 dx 3 [ f(x)] ⋮ ⋮ These are called, in succession, the first derivative, the second derivative, the third derivative, and so forth. The number of times that f is differentiated is called the order of the derivative.

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