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Welcome To Calculus (Do not be afraid, for I am with you) The slope of a tangent line

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Questions To Think About If you drive 150 miles in 3 hours, what’s your average speed? –50 mph is your AVERAGE RATE OF CHANGE What is the general maths term for AVERAGE RATE OF CHANGE? –SLOPE of a line = AVERAGE RATE OF CHANGE Did you really drive 50 mph constantly on your journey? –NO…that was your AVERAGE RATE OF CHANGE Fill in the blank: When driving, I looked at my speedometer and it read 65 mph. At this instant, 65 mph was my _______________________ rate of change. INSTANTANEOUS

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150 Distance In Miles Time In Hours 3 Tell me something about your INSTANTANEOUS RATE OF CHANGE 1 / 2 hour into the trip. Pretty Fast (Got stopped by the police about 15 minutes later) Tell me something about your INSTANTANEOUS RATE OF CHANGE 2 hours into the trip. Went backwards because I was lost miles in 3 hours = AVE RATE of 50 mph

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Let’s find the slope of the tangent line to y = x 2 when x = 2 (Instantaneous Rate of Change) The Slope Of The Red Line

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We begin by setting up what’s called a secant line through (2,4) and (2+h,(2+h) 2 ) (2+h,(2+h) 2 ) (2,4) h (2+h) The slope of that line =

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As h gets smaller, the secant line approaches the tangent line, and the average rate of change becomes the instantaneous rate of change h h As h gets closer and closer to zero, we approach our tangent line

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When h approaches zero, our slope equation becomes…. (2+h,(2+h) 2 ) (2,4) h (2+h) The slope of tangent line =

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Lets Evaluate The Limit == = == = 4 The Slope of the Tangent Line at (2,4) = 4

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(x+h,f(x+h)) (x,f(x)) (x+h,f(x+h)) · y = f(x) CLICK TO CONTINUE

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Find the slope of the tangent line for any point (x,f(x)) for f(x)=x 2 · · Start with the slope of a secant

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Now lets make it a tangent ·

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The formula that will calculate the slope of a tangent line for at any point (x,y) is 2x For example: At (3,9) the slope of the tangent is … 6 At (-4,16) the slope of the tangent is … -8 We say that the derivative of is 2x

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SUMMARY To find the equation of the tangent line, simply find f ’(a), that is your slope. Now use your point of tangency {(a,f(a)} and your slope, m = f ‘(a) in the point – slope form of a linear equation.

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