Section 2.7- Derivative and Rate of Change

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1 Section 2.7 Derivatives and Rates of change Recall the tangent line problem in section 2.1 We define the slope of the tangent line at P to be the limit: ? ?? = lim ğ‘¥â†’ğ‘Ž ?(?) − ?(ğ‘Ž) ? − ğ‘Ž Ex1: Find the equation of the tangent line to ? = ? 2 ğ‘Ž? 𝑃(1,1) There is another expression for the slope of a tangent line that is sometimes easier to use. If we let ℎ = ? − ğ‘Ž ?ℎ?? ? = ğ‘Ž + ℎ . Also, ? → ğ‘Ž ??ğ‘Ž?? ℎ → 0 ? ?? =
2 Ex2: Find the equation of the tangent line to the hyperbola ? = 3 𝑥 ğ‘Ž? ? = 3 Velocity: In general, suppose an object moves along a straight line according to an equation of motion s = f(t) , where ? is the displacement (directed distance) of the object from the origin at time ? . The function ? that describes the motion is called the position function of the object. In the time interval from ? = ğ‘Ž to ? = ğ‘Ž + ℎ the change in position is ?(ğ‘Ž + ℎ) − ?(ğ‘Ž) average velocity = displacement time elapsed = f(a+h)−f(a) h Instantaneous velocity is the limit of average velocities: 𝒗(𝒂) = ?𝐢? ğ’‰â†’ğŸŽ 𝒇(𝒂+𝒉)−𝒇(𝒂) 𝒉 ( Instantaneous velocity at 𝒕 = 𝒂 ) (Speed: is the absolute value of velocity) Note: if v >0 means the particle moves to the positive (right or up) direction If v<0 means the particle moves to the negative (left or down) direction If v=0 means the particle does not change its position.
3 Ex3: Suppose that a ball is dropped from the upper observation deck of a tower 450 m above the ground. (a) What is the velocity of the ball after 5 seconds? (b) How fast is the ball traveling when it hits the ground? Solution: We will need the velocity both when t = 5 and when the ball hits the ground. So it is efficient to find the velocity at a general time t, then evaluate it at the needed values. Recall from 2.1: The distance (in meters) fallen after t seconds is ?(?) = ?(?) = 4.9? 2 Now: instantaneous velocity at time t : 𝑣(?) = ?𝐢? ğ’‰â†’ğŸŽ 𝒇(𝒕+𝒉)−𝒇(𝒕) 𝒉
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