Pg 869 #1, 5, 9, 12, 13, 15, 16, 18, 20, 21, 23, 25, 30, 32, 34, 35, 37, 40, 42 16.3 Tangent to a Curve.

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Presentation transcript:

Pg 869 #1, 5, 9, 12, 13, 15, 16, 18, 20, 21, 23, 25, 30, 32, 34, 35, 37, 40, 42 16.3 Tangent to a Curve

What if you were asked to find the slope of a curve? Could you do this? Does it make sense? (No, not really, slopes are for lines, they are straight, curves might not be straight) So, what if I told you this is exactly what we are going to do! We will be utilizing limits!! Let’s think about geometry for a second. A secant was a line that intersected a circle at two points. A tangent was a line that intersected a circle at just one point. Let’s extend this to a curve – any curve! Secant Line to a Curve on Desmos (Slide the dot on the right slowly towards the dot on the left) The secant line becomes a tangent line!

So, we want to make the secant become the tangent. What is the slope of Q (x, f (x)) (c, f (c)) P We want Q to get closer to P So x needs to get closer to c c x this is the difference quotient!!

Ex 1) Find the slope of the line tangent to the curve at P(5, 3) any other point (x, f (x)) (5, 3) Draw your own sketch slope We can’t substitute 5 in, so algebra to work! *now we plug in 5*

*to find equation of a line, we need two things: (1) slope (2) point If we know the slope of the tangent line, we can write the equation of the tangent line. Ex 2) Find equation of tangent line to f (x) = x3 – 2x2 + 3 at P(1, 2). *to find equation of a line, we need two things: (1) slope (2) point *you can leave like this – that is what calculus does m = –1 P(1, 2) y – 2 = –1(x – 1) So back to an original question – how to find slope of a curve… the slope of a curve at point P the slope of the tangent at point P The slope of a curve might vary from point to point, so it is helpful to be able to represent it in generic form using an arbitrary point. Then we can use it with specific slope values.

Ex 3) Find equation of line with slope 5 tangent to the graph of f (x) = x2 + 3x – 1 This time we have slope, but not the point general terms: f (x) and f (c) want 2c + 3 = 5 2c = 2 c = 1 point (1, ? ) f (1) = 1 + 3 – 1 = 3 (1, 3) y – 3 = 5(x – 1)

Average rates are similar to secants (slope of line) Physical quantities can also be found using the idea of a secant becoming a tangent. Average rates are similar to secants (slope of line) Instantaneous rates are similar to tangents (limit of slope of line) Let’s look at velocity (a rate!) A position function f (t) describes the path something takes. Ex 4) The motion of an object is given by the function f (t) = t2 – 3t + 5 where f (t) is height of object in feet at time t seconds. What is the average velocity of the object between t = 2 s and t = 4 s? slope of secant b) What is the instantaneous velocity of the object at time t = 2 s? slope of tangent f (t) & f (2)

Homework Pg 869 #1, 5, 9, 12, 13, 15, 16, 18, 20, 21, 23, 25, 30, 32, 34, 35, 37, 40, 42