Tangent Lines (Sections 2.1 and 3.1 )

Slides:



Advertisements
Similar presentations
Unit 6 – Fundamentals of Calculus Section 6
Advertisements

Tangent Lines ( Sections 1.4 and 2.1 ) Alex Karassev.
Find the slope of the tangent line to the graph of f at the point ( - 1, 10 ). f ( x ) = 6 - 4x
Calculus 2413 Ch 3 Section 1 Slope, Tangent Lines, and Derivatives.
Derivatives - Equation of the Tangent Line Now that we can find the slope of the tangent line of a function at a given point, we need to find the equation.
DO NOW: Use Composite of Continuous Functions THM to show f(x) is continuous.
Rates of Change and Tangent Lines Section 2.4. Average Rates of Change The average rate of change of a quantity over a period of time is the amount of.
1.4 – Differentiation Using Limits of Difference Quotients
2.4 Rates of Change and Tangent Lines
Section 6.1: Euler’s Method. Local Linearity and Differential Equations Slope at (2,0): Tangent line at (2,0): Not a good approximation. Consider smaller.
Find an equation of the tangent line to the curve at the point (2,3)
More Review for Test I. Calculation of Limits Continuity Analytic Definition: the function f is continuous at x = a if When you can calculate limits.
1.6 – Tangent Lines and Slopes Slope of Secant Line Slope of Tangent Line Equation of Tangent Line Equation of Normal Line Slope of Tangent =
Assignment 4 Section 3.1 The Derivative and Tangent Line Problem.
Section 2.6 Tangents, Velocities and Other Rates of Change AP Calculus September 18, 2009 Berkley High School, D2B2.
Dr. Omar Al Jadaan Assistant Professor – Computer Science & Mathematics General Education Department Mathematics.
Tangents. The slope of the secant line is given by The tangent line’s slope at point a is given by ax.
Directional Derivatives and Gradients
Linear approximation and differentials (Section 2.9)
Section 5.5 The Intermediate Value Theorem Rolle’s Theorem The Mean Value Theorem 3.6.
Solve and show work! State Standard – 4.1 Students demonstrate an understanding of the derivative of a function as the slope of the tangent line.
Review: 1) What is a tangent line? 2) What is a secant line? 3) What is a normal line?
2.1 The Derivative and The Tangent Line Problem Slope of a Tangent Line.
Local Linear Approximation Objective: To estimate values using a local linear approximation.
Newton’s Method Problem: need to solve an equation of the form f(x)=0. Graphically, the solutions correspond to the points of intersection of the.
Section 9-7 Circles and Lengths of Segments. Theorem 9-11 When two chords intersect inside a circle, the product of the segments of one chord equals the.
1 10 X 8/30/10 8/ XX X 3 Warm up p.45 #1, 3, 50 p.45 #1, 3, 50.
2.1 The Derivative and the Tangent Line Problem Objectives: -Students will find the slope of the tangent line to a curve at a point -Students will use.
Unit 2 Lesson #3 Tangent Line Problems
Lecture 5 Difference Quotients and Derivatives. f ‘ (a) = slope of tangent at (a, f(a)) Should be “best approximating line to the graph at the point (a,f(a))”
The Derivative and the Tangent Line Problem Section 2.1.
Calculus Section 3.1 Calculate the derivative of a function using the limit definition Recall: The slope of a line is given by the formula m = y 2 – y.
Calculating Derivatives From first principles!. 2.1 The Derivative as a Limit See the gsp demo demodemo Let P be any point on the graph of the function.
3.2 Rolle’s Theorem and the
Rolle’s theorem and Mean Value Theorem (Section 4.2)
Mean Value Theorem.
Linear approximation and differentials (Section 3.9)
The gradient is 0.
Chapter 16A.
2.1 Tangents & Velocities.
Section 11.3A Introduction to Derivatives
§ 1.2 The Slope of a Curve at a Point.
Rate of Change.
Find the equation of the tangent line to the curve y = 1 / x that is parallel to the secant line which runs through the points on the curve with x - coordinates.
Find the equation of the tangent line for y = x2 + 6 at x = 3
Analysing a function near a point on its graph.
2.1A Tangent Lines & Derivatives
2.7 Derivatives and Rates of Change
Polynomials: Graphing Polynomials. By Mr Porter.
2.4 Rates of Change & Tangent Lines
The Tangent and Velocity Problems
3.2 Rolle’s Theorem and the
Derivatives by Definition
Question Find the derivative of Sol..
F’ means derivative or in this case slope.
Section 2.7.
Point-slope Form of Equations of Straight Lines
2-4: Writing Linear Equations Using Slope Intercept Form
30 – Instantaneous Rate of Change No Calculator
that ordered pair is the one solution.
The derivative as the slope of the tangent line
Linear approximation and differentials (Section 3.9)
Point-slope Form of Equations of Straight Lines
CPE 332 Computer Engineering Mathematics II
Slope Fields (6.1) January 10th, 2017.
Sec 2.7: Derivative and Rates of Change
Linear Approximation.
Chapter 3 Additional Derivative Topics
2.4 The Derivative.
Graphing Systems of Equations.
Presentation transcript:

Tangent Lines (Sections 2.1 and 3.1 ) Alex Karassev

Tangent line ? What is a tangent line to a curve on the plane? Simple case: for a circle, a line that has only one common point with the circle is called tangent line to the circle This does not work in general! ? P P

Idea: approximate tangent line by secant lines Secant line intersects the curve at the point P and some other point, Px y y Px Px P P x x a x x a

Tangent line as the limit of secant lines Suppose the first coordinate of the point P is a As x → a, Px x → a, and the secant line approaches a limiting position, which we will call the tangent line y y Px Px P P x x a x x a

Slope of the tangent line Since the tangent line is the limit of secant lines, slope of the tangent line is the limit of slopes of secant lines P has coordinates (a,f(a)) Px has coordinates (x,f(x)) Secant line is the line through P and Px Thus the slope of secant line is: m y y=f(x) mx P Px f(x) x x a

Slope of the tangent line We define slope m of the tangent line as the limit of slopes of secant lines as x approaches a: Thus we have: m y y=f(x) mx P Px f(x) x x a

Example Find equation of the tangent line to curve y=x2 at the point (2,4) y P Px x 2

Solution We already know that the point (2,4) is on the tangent line, so we need to find the slope of the tangent line P has coordinates (2,4) Px has coordinates (x,x2) Thus the slope of secant line is: y P Px x 2

Solution Now we compute the slope of the tangent line by computing the limit as x approaches 2: y P Px x 2

Solution y Thus the slope of tangent line is 4 and therefore the equation of the tangent line is y – 4 = 4 (x – 2) , or equivalently y = 4x – 4 P Px x 2