Falling Cat 2. Falling Cat 3. Falling Cats 5. Falling Cat 4. Acceleration due to Gravity Consider a cat falling from a branch

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1 Calculus for the Life Sciences Lecture Notes Velocit and Tangent Joseph M. Mahaff, Department of Mathematics and Statistics Dnamical Sstems Group Computational Sciences Research Center San Diego State Universit San Diego, CA jmahaff Outline Spring 7 (/47) (/47) Falling Cat Objects falling under the influence of gravit are important in classical differential Calculus Sir Isaac Newton s work on gravit was a ke step to the development of Calculus Controvers as to whether Newton or Gottfried Leibnitz was the first to invent Calculus Cat have evolved to be one of the best mammalian predators Domestic cats have been shown to responsible for up to 6% of the deaths of songbirds in some communities The are adapted to hunting in trees Cats have a ver fleible spine for hunting This fleibilit allows them to rotate rapidl during a fall (3/47) (4/47)

2 Falling Cat Falling Cat 3 Humans are fascinated b this abilit of a cat to right itself Jared Diamond Stud of Cats falling out of New York apartments Paradoicall the cats falling from the highest apartments actuall fared better than ones falling from an intermediate height The cat remains tense earl in the fall With greater heights the falling cat relaes and spreads its legs to form a parachute This slows its velocit a little and results in a more even impact From intermediate heights, the cat basicall achieves terminal velocit, but the tension causes increased likelihood of severe or fatal injuries (5/47) Acceleration due to Gravit Consider a cat falling from a branch The earl stages of the fall result from acceleration due to gravit Newton s law of motion sas that mass times acceleration is equal to the sum of all the forces acting on an object Velocit is the derivative of position Acceleration is the derivative of velocit (6/47) Falling Cat 4 5 Suppose that a cat falls from a branch that is 6 feet high The height of the cat satisfies the equation How long does this cat fall? h(t) = 6 6t What is its velocit when it hits the ground? From the equation, the cat hits the ground when h(t) = 6 6t = This occurs when t = However, the velocit at t = requires more work We will show that the cat has a velocit, v() = 3 ft/sec (about.8 mph) (7/47) (8/47)

3 of the Falling Cat Suppose that the height of an object is given b h(t) The between times t and t satisfies: v ave = h(t ) h(t ) t t. Return to the cat falling from a 6 ft tree limb, where h(t) = 6 6t Consider the average velocit of the falling cat between t = and t = : v ave = h() h(.5) = = 4ft/sec..5.5 Consider the average velocit of the falling cat between t =.9 and t = : v ave = h() h(.9) = 3.4 = 3.4ft/sec..9. Consider the average velocit of the falling cat between t =.99 and t = : v ave = h() h(.99).99 =.384. = 3.84ft/sec. (9/47) (/47) Velocit of the Falling Cat Return to the cat falling from a 6 ft tree limb, where h(t) = 6 6t Recall the cat hits the ground at t = sec We find the general secant line between t = z and t =, which relates to the near t = Since h( z) = 6 6( z) = 3z 6z v ave = h() h( z) ( z) = 3z +6z z = 3+6z Consider a ball thrown verticall under the influence of gravit, ignoring air resistance The ball begins at ground level (h() = cm) It is thrown verticall with an initial velocit, v() = 96 cm/sec The acceleration of gravit is g = 98 cm/sec The height of the ball for an time t ( t 4) is given b h(t) = 96t 49t As z, v ave 3, so the cat hits the ground at a velocit of 3 ft/sec (.8 mph) (/47) (/47)

4 3 Graph of the height of a ball for t 4, showing position ever.5 sec Height (cm) 5 5 Height of Ball 3 4 Time (sec) Compute the average velocit between each point on the graph The average velocit is the difference between the heights at two times divided b the length of the time period Associate the average velocit with the midpoint between each time interval Height (t ) Height (t ) Average Time h(t ) h(t ) t a = (t +t )/ v(t a ) = h(t ) h(t ) (t t ) h() = h(.5) = t a =.5/ =.5 v(.5) = 75 h(.5) = h() = 96 t a =.75 v(.75) = 45 h(3) = 47 h(3.5) = t a = 3.5 v(3.5) = 5 (3/47) (4/47) 4 Flight of Ball 5 Graph of the velocit of a ball for t 4, showing velocit ever.5 sec Velocit (cm/sec) of Ball Time (sec) The graph of the height of the ball as a function of time is a parabola The graph of the velocit of the ball as a function of time is a line The average velocit is zero when the ball reaches its maimum height The verte of the parabola (maimum height of the ball) is where the velocit is zero (t-intercept) (5/47) (6/47)

5 6 7 Graph of the height of a ball for t 4, showing position ever. sec Height (cm) Height of Ball 5 How does this affect the average velocit computation? The distance between successive heights is now closer But then the intervening time interval is also closer together The average velocit between t =. and t =.3 has h(t ) = 37.4 cm and h(t ) = cm, so v(.5) = 75 cm/sec, the same as before Time (sec) (7/47) (8/47) Flight of Ball 8 Flight of Ball 9 Graph of the velocit of a ball for t 4, showing velocit ever. sec Velocit (cm/sec) of Ball Time (sec) The average velocit data lie on the same straight line as before v(t) = 96 98t This straight line function is the derivative of the quadratic height function h(t) The calculation suggests that the derivative function is independent of the length of the time interval chosen This is specific to the quadratic nature of the height function Soon we will learn to take derivatives of more functions (9/47) (/47)

6 Eample Leaping Salmon Eample Leaping Salmon A river is dammed, and a salmon ladder is built to enable the salmon to bpass the dam and continue to travel upstream to spawn The vertical walls on the salmon ladder are 6 feet high The salmon has to leap verticall upwards over the wall The height of the salmon during its leap is given b Skip Eample h(t) = v t 6t Let v = ft/sec. Sketch a graph of the height of the salmon h(t), with time, showing clearl the maimum height and when the salmon can clear the wall Find the average velocit of the salmon between t = and t =.5 and associate this velocit with t =.5 Repeat this process for each half-second of the leaping salmon, then sketch a graph of the average velocit as a function of time, t Determine the minimum speed, v, that the salmon needs on eiting the water to climb the salmon ladder (/47) (/47) Eample Leaping Salmon 3 Solution: The function h(t) is a parabola, h(t) = t 6t = 4t(5 4t) The t-intercepts are t = and t =.5 The verte occurs at (.65, 6.5) The salmon can clear the wall when h(t) = 6, so t 6t = 6 or 8t t+3 = This can be factored to give (t )(4t 3) = The salmon can clear the wall at an time < t < 3 4 sec (3/47) Eample Leaping Salmon 4 Graph of h(t) = t 6t h(t) (ft) Leaping Salmon (.65,6.5) (.5,6) t (sec) (4/47)

7 Eample Leaping Salmon 5 Solution (cont): The average velocit of the salmon between t = and t =.5 is given b, v(.5) = h(.5) h().5 = ((.5) 6(.5) ).5 The average velocit of the salmon between t =.5 and t = is given b v(.75) = h() h(.5).5 = = 4 ft/sec = ft/sec Eample Leaping Salmon 6 Graph of average velocit of the salmon satisfing Velocit (ft/sec) v ave (t) = 3t (.5,) Salmon Velocit (.75, 4) (5/47) t (sec) (6/47) Eample Leaping Salmon 7 Solution (cont): The minimum speed, v, that the salmon needs to climb the fish ladder is the one that produces a maimum height of 6 ft h(t) = v t 6t The t value of the verte occurs at t = v ( 6) = v 3 Since we want the verte to be 6 ft, ( v ) ( v ) ( v ) v h = v 6 = = 6. Secant Lines and Tangent Line The average velocit is the same calculation as the slope between the two data points of the height function The slope of the secant line between two points on a curve Geometricall, as the points on the curve get closer together, then the secant line approaches the tangent line The tangent line represents the best linear approimation to the curve near a given point Its slope is the derivative of the function at that point v = ft/sec (7/47) (8/47)

8 Secant Lines and Tangent Line Secant Lines and Tangent Line Definition: A secant line for a curve is a line that connect two points on the curve. Graph showing 5 Tangent Line 4 Definition: A tangent line for a curve is a line that touches the curve at eactl one point and provides the best approimation to the curve at that point. 3 Secant Line (9/47) (3/47) Tangent Line Eample = A tangent line represents the best linear approimation to the curve near a given point Consider the function = Tangent Line = f() Tangent Line = f() = f() Tangent Line Find the equation of the tangent line at the point (,) on the graph A secant line is found b taking two points on the curve and finding the equation of the line through those points Create a sequence of secant lines that converge to the tangent line b taking the two points closer and closer together (3/47) (3/47)

9 Eample = Consider the secant line through the points (,) and (,4) This line has a slope of m = 3, and its equation is = 3 Consider the pair of points on the curve =, (,) and (.5,.5) This line has a slope of m =.5, and its equation is =.5.5 The secant line through the points (,) and (.,.) has a slope of m =. Its equation is =.. Eample = 3 Graph of = with secant lines = = 3 =.5.5 =.. = (,) (, 4) (.,.) (.5,.5) (33/47) (34/47) Eample = 4 General secant line for = at (,) Consider the value = +h for some small h The corresponding value = (+h) = +h+h The slope of the secant line through this point and the point (,) is m = (+h+h ) (+h) The formula for this secant line is = h+h h = (+h) (+h) = +h Eample = 5 The general secant line for = through (,) is = (+h) (+h) As h gets ver small, the secant line gets ver close to the tangent line Its not hard to see that the tangent line for = at (,) is = The slope of the tangent line is m = The value of the derivative of = at = (35/47) (36/47)

10 The geometric view of the tangent line is ver eas to visualize The graph on the left is f() with tangent lines shown, while the graph on the right is the derivative of f() 6 4 Tangent Lines for Cubic Equation Slope of Tangent Lines for Cubic (37/47) Several points of interest The graph on the left is a cubic function, while the graph of its derivative is a quadratic As ou approach a maimum (or minimum) for the cubic function, the value of the derivative goes to zero and the sign of the derivative function changes This is an important application of the derivative (38/47) Eample Secant Lines Consider the function Skip Eample f() = Let all secant lines have the point, =. Other points of the sequence have =, =.5, =., =., and =. Find the derivative of f() at = b finding the slope of the tangent line at = Graph f(), the tangent line, and the secant lines Eample Secant Lines Solution: This eample eamines secant lines for through the point (,) f() = When =, f() =, so the secant line has slope m = and is given b = For =.5, two points on the secant line are (,) and (.5,.75), which gives the secant line =.5.5 (39/47) (4/47)

11 Eample Secant Lines 3 Solution (cont): Continuing the process: When =., two points on the secant line are (,) and (.,.4), which gives the secant line =.. For =., two points on the secant line are (,) and (.,.), which gives the secant line =.. For =., two points on the secant line are (,) and (.,.), which gives the secant line =.. (4/47) Eample Secant Lines 4 Solution (cont): The pattern in the sequence easil gives the tangent line = 4 3 = = =.5.5 =.. =.. = f() = (4/47) Eample Secant Lines 5 Eample Secant Lines 6 Solution (cont): Let s find the slope of the secant line through the points Solution (cont): Since the tangent line has slope m =, the derivative of f() = at = is Since patterns cannot alwas be recognizable, we need a better wa to compute the derivative (,f()) = (,) and (+h,f(+h)) Since f(+h) = (+h) (+h) = h +h, the slope of the secant line is m(h) = (h +h) (+h) = h +h h = +h (43/47) As h, m(h) It follows that the slope of the tangent line is, which is the derivative of f() at = (44/47)

12 Eample Square Root Function Consider the function f() = + Find the slope of the secant line through the points (,f()) and (+h,f(+h)) Let h get small and determine the slope of the tangent line through (,), which gives the value of the derivative of f() at = Eample Square Root Function Solution: The slope of the secant line is m(h) = f(+h) f() (+h) +h+ + 4+h = = h h ( )( ) 4+h 4+h+ = h 4+h+ = = 4+h 4 h( 4+h+) 4+h+ (45/47) (46/47) Eample Square Root Function 3 Solution (cont): The slope of the secant line is m(h) = 4+h+ In the formula above, as h, the slope of secant line, m, approaches m t = = 4+ 4 Since the derivative is related to the limiting case of the slope of the secant lines (the slope of the tangent line, m t ), we see that the derivative of f() at = must be 4 (47/47)