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Preview of Calculus

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- 1. Section 1.1 A Preview of Calculus What is it good for, anyway? Look at p. 43 - 44 for some examples
- 2. Calculus makes the concepts of precalculus DYNAMIC! PRECALCULUS LIMITING PROCESS CALCULUS
- 3. There are two big questions in Calculus! What is the equation of the tangent line that touches a curve at a particular point? What is the area underneath a curve?
- 4. To get the equation of a tangent line, we must know a point and a slope. How do we find the slope at a given point?
- 5. We can find slope if we know two points!
- 6. We can define slope for lines, how can that carry over to curves? <ul><li>We can approximate the slope of a curve at a given point by finding the slope of a secant line using that point and one close by. </li></ul><ul><li>Look at figure 1.2 on p. 45 in book. </li></ul>
- 7. P(c, f(c)) is the point of tangency. Q(c+ Δ x, f(c+ Δ x)) is nearby point
- 8. What is the slope of the secant line? or
- 9. Here is a great site to see what is happening! <ul><li>http://www.math.psu.edu/dlittle/java/calculus/secantlines.html </li></ul>
- 10. <ul><li>What did you come up with for slope? </li></ul><ul><li>What is the point of tangency? </li></ul><ul><li>What is the equation of the tangent line? Hint: Remember point slope formula for an equation of a line: Given slope m and point (x 1 , y 1 ): </li></ul><ul><li>(y – y 1 ) = m(x – x 1 ) </li></ul>
- 11. <ul><li>If you got even closer to the point by letting Δ x get really small, what would happen? </li></ul>
- 12. How do you find area under a curve? Fig. 1.3, p.46
- 13. Let’s divide it into rectangles that approximate the area!
- 14. So, how wide are the rectangles? How tall are they? If we divide into 4 rectangles, the width would be The height would be the y-value at the left or right side of rectangle, or interval.
- 15. Rect. 1 Rect 2 Rect 3 Rect 4 Width .25 .25 .25 .25 Height f( )= f( )= f( )= f( )= Area = W ٠ H
- 16. If rectangles formed from right of interval: .25 ٠ 0 + .25 ٠ .0625 + .25 ٠ .25 + .25 ٠ .5625 = 0 + .01563 + .0625 + .14063 = .21876 If rectangles formed from left of interval: .25 ٠ .0625 + .25 ٠ .25 + .25 ٠ .5625 + .25 ٠ 1 = .01563 + .0625 + .14063 + .25 = .46876 The first is an underestimate, the second is an overestimate. The actual value is 0.3333…
- 17. What would happen if we increased the number of rectangles? http://www.math.psu.edu/dlittle/java/calculus/area.html p. 47/ 1-9 odd

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