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Section 2.6 Slopes of tangents SWBAT: Find slopes of tangent lines Calculate velocities

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Review of Secant lines Remember this!?! the slope of the secant line PQ :the slope of the secant line PQ :

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Concept: take the limit x approaches a.

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Slope of a Tangent to f(x) at point ( a, f(a) ) is:

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Example Start with y = x 2 at the point P(2, 4). Use our definition of the slope of the tangent at a point. Need a hint:

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Example (cont’d)

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Using the point-slope form of the equation of a line y – 4 = 4(x – 2) So,y = 4x – 4

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Another Expression (cont’d)

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Another Expression There is another way to define slope of a tangent: We let h = x – aWe let h = x – a Then x = a + hThen x = a + h Thus the slope of secant line PQ isThus the slope of secant line PQ is

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A second definition is:

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Example 2: Use to calculate the slope of f(x) at a.

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Example 3. Find the equation of the tangent line to the hyperbola y = 3/x at the point (3, 1)

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Example (cont’d)

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Solution Let f(x) = 3/x. Then the slope of the tangent at (3, 1) is – ⅓. Therefore an equation of the tangent line is y – 1 = – ⅓ (x – 3), or y= ⅓ x+2.

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Example (cont’d)

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Velocity The average velocity over a time interval h equals the slope of the secant line PQ.

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Velocity (cont’d)

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Taking it to the limit we get... Instantaneous Velocity This means that the velocity at time t = a is equal to thevelocity at time t = a is equal to the slope of the tangent line at P.slope of the tangent line at P.

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Example (Calculator active) Suppose a ball is dropped from the top of a tower 450m. Tall. Find the velocity at 5 seconds. Position function f(t) = 4.9t 2

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Example (cont’d) a) The velocity after 5 s is v(t)=9.8t v(5) = (9.8)(5) = 49 m/s.

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Part deux: How fast is the ball going when it hits the ground?

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Example (cont’d) It hits the ground when it traveled 450m The ball will hit the ground when, 4.9t 2 = 450. Solving for t gives t ≈ 9.6 s. The velocity of the ball as it hits the ground is therefore v(t) = 9.8t ≈ 94 m/s

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Assignment 11 P. 145 1-13 odd

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Other Rates of Change In general, we can have other average rates of change: These can lead to instantaneous rates of change.

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Other Rates (cont’d)

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Example Temperature readings T (in °C) were recorded every hour starting at midnight on a day in Whitefish, Montana. The time x is measured in hours from midnight. The data are given in the table on the next slide:

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Example (cont’d)

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a) Find the average rate of change of temperature with respect to time i.from noon to 3 P.M. ii.from noon to 2 P.M. iii.from noon to 1 P.M. b) Estimate the instantaneous rate of change at noon.

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Solution to (a) i. From noon to 3 P.M. the temperature changes from 14.3 °C to 18.2 °C, so ∆T = T(15) – T(12) = 18.2 – 14.3 = 3.9 °C while the change in time is ∆x = 3 h. Therefore, the average rate of change of temperature with respect to time is

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Solution to (a) (cont’d) ii. From noon to 2 P.M. the average rate of change is iii. From noon to 1 P.M. the average rate of change is

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Solution to (b) We plot the given data on the next slide and sketch a smooth curve of the temperature function. Then we draw the tangent at the point P where x = 12. By measuring the sides of triangle ABC, we estimate the slope of the tangent line to be 10.3/5.5 ≈ 1.9.

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Solution to (b) (cont’d)

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Therefore the instantaneous rate of change of temperature with respect to time at noon is about 1.9 °C.

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Review Use of limits to define Tangent linesTangent lines Instantaneous velocitiesInstantaneous velocities Other rates of changeOther rates of change

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The derivative of a function f(x), denoted f’(x) is the slope of a tangent line to a curve at any given point. Or the slope of a curve at any given.

The derivative of a function f(x), denoted f’(x) is the slope of a tangent line to a curve at any given point. Or the slope of a curve at any given.

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