Calculus in the AP Physics C Course The Derivative

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1 Limits an Derivatives Calculus in the AP Physics C Course The Derivative In physics, the ieas of the rate change of a quantity (along with the slope of a tangent line) an the area uner a curve are essential. Limits are funamental for the efinitions of the two major concepts in calculus that escribe these ieas: the erivative an the integral Our intent here is to introuce these ieas to AP Physics C stuents who have most likely not yet covere them in a calculus course. Mathematical rigor has been left for the AP Calculus teacher to cover an a more intuitive approach is use here. Limits Limits are concerne with etermining the values of functions base on their behavior near a value a. Often stuents are incline to think that the value of a function etermine from its behavior near = a is eactly the same as the value of the function at a. In teaching the notion of the limit we must make the istinction between behavior near a an value at a. 4 Consier Graph 1: f ( 1 If we begin taking values to evaluate f (, we see that as approaches, f ( approaches 5. Note that there is no f (), so we can't write f ( ) 5. But as we get closer an closer to (i.e., ), f ( write lim f ( 5 gets closer an closer to the limit 5. We can 1

2 Consier Graph : f ( 4. 5 Note that there is no f (). This time we nee to approach from both sies. Approaching from the left (-): Approaching from the right (+): lim f ( 4 lim f ( 5 Again, we can get closer an closer to, but we cannot compute f () irectly, an can only approach the value of f ( near by using the concept of the limit. Infinite Limits: Graph 1 f ( Note that the function is asymptotic to, an there is no f ().

3 Approaching from the right (+): Approaching from the left (-): There is no value for We o not write Limits as lim f ( ( f ( gets unbounely large) lim f ( ( f ( gets unbounely large an negative) lim f ( since the left an right limits on't agree. lim f (. 4 : Graph 4 f ( 4 1 Note that the limit of a function may be a particular value even if f never reaches that value. The limit must be approache, but not necessarily attaine. We have lim f ( 4, although f ( never attains 4. Now for functions which are continuous at a particular point: Theorem: If f is continuous at = a, so that its graph oes not break, then For eample, recall Graph : lim f ( f ( 1) 4 1 lim a f ( f ( a) Another eample: Consier the continuous function f ( As, lim f ( lim( () 4. Which simply means f ( ) 4 Now, on to the erivative get ecite

4 The Derivative as the Slope of a Tangent The erivative of a function represents the rate of change of the function with respect to another quantity. The erivative also represents the slope of the line tangent to the graph of a function at a particular point. Consier motion at a constant spee. The position,, vs time, t, graph of this motion looks like this: But what if the motion is accelerate? The spee (an thus slope) woul be continually changing. We can approimate the spee at a particular point P by rawing a line tangent to the point an calculating its slope. P slope of tangent line = antaneous spee at point P But how o we fin the slope of a line tangent to a particular point on a curve? Choose a point Q on the curve near point P an connect the two points with a line calle the secant line, as shown below. The slope of the secant line is the average spee between points P an Q, an is approimately equal to the antaneous spee at point P. 4

5 If we allow point Q to approach point P, our ' s an t' s will become smaller an smaller (a limiting process as t 0 ) the secant line will become the tangent line at P, an its slope will represent the antaneous spee at point P. The slope of a line tangent to a point P on a curve is calle the erivative of the function escribing the curve. We will return to this relationship later. (Δt getting smaller) The Derivative as the Rate of Change of a Function Suppose that the position of a car on a roa at any time t is t 1, f ( 1) 11, at t, f ( ) 16, at t, f ( ) 9, an so on. f ( t) 1t t so that at: What is the speeometer reaing v at any time t? We know that average spee total change in position v total change in time Consier the perio between times t an, t t. The average velocity between these two times is change in position v change in time later position earlier position change in time But how o we fin the antaneous velocity? Take smaller an smaller time intervals approach zero an take the limit. After epaning, we fin that: t, that is, allow t to v (t t) (t) 1(t t) (t t) (1t t ) lim = lim t0 t t0 t 1t 1t (t t t tt t ) 1t t 1t t = lim = lim t0 t t0 t tt t t 5

6 But, since we are taking the limit ast 0, we are left with v lim(1 t tt t ) = 1 t t0 The function, 1 t is the erivative of f with respect to t an is enote f '( t). f ( t) 1t t v f '( t) 1 t (rate of change of with respect to t) (a) If the erivative (antaneous velocity) is positive at a certain time, the car is moving to the right at that time. (b) If the erivative is negative at a certain time the car is moving to the left at that time. (c) What oes it mean if the erivative at a certain time is zero? () What is this car's spee at t=0? We have seen that a erivative is the rate of change of a function an the slope of a line tangent to a point on a curve. Let's generalize the slope of a line tangent to a particular point on the curve. As one earlier we choose another point Q at the point [, f ( ], an raw the secant line through P an Q. We now allow the secant line to become the tangent line at P using the limiting process as 0. As before, in terms of limits, erivative f ( f '( lim 0 f ( Then f '( change in y y lim lim change in 0 0 y The erivative of y with respect to can be written several ifferent ways, but usually, we will write y y erivative of y with respect to. Keep in min that is not a quotient (although we can sometimes treat it as one), but is only our notation for a erivative, our antaneous rate of change. Fining the erivative the easy way: Recall from our previous eample that f ( t) 1t t an v t 1 t In general, the erivative of Eample 1. n n n1 This is calle the POWER RULE an it will get us pretty far in the application of ifferential calculus to physics. y

7 Then, taking the erivative of y with respect to, we write: y Note that the erivative of a constant (in this case, 6) is always zero, since (a) a constant by efinition has no rate of change, (b) graphically, the slope of a constant function is zero 0 01 (c) we coul write 6 as 6, an its erivative woul be (0)(6 ) 0 Eample : Consier the position as a function of time: 8t t 1 (a) At t 0, what is the position of the object? (b) At This is pretty simple: we just substitute 0 in for t: 8(0) (0) 1 = -1 (We really shoul put some units most likely meters in here, but I on t want to complicate this problem) t s, what is the velocity of the object? This is a little more ifficult, but we just emonstrate that the antaneous velocity at any time t can be foun by fining the erivative for the position vs. time function. f (t) 8t v t f '(t) 4t 1 6t By substituting t = secons into the equation for v, we get v f '() 4() 6() 84 (c) When is the velocity of the object zero? If you think about what this question is asking, you ll see that solving this requires us to use the equation we have for the antaneous velocity, v equal to zero: v f '(t) 4t 6t 0 The roots of this equation are t = 0, an 4 secons. () Is the object accelerating? Simply looking at the equation for v, we see that the velocity varies with time. Therefore, accoring to our efinition for acceleration, the object is accelerating. (e) Can we fin its acceleration? We know that the slope of a velocity vs. time graph represents acceleration. That means, if we take the erivative of the velocity vs. time function, we will get an acceleration vs. time function. We sometimes call this the secon erivative of the position vs. time function: v f '(t) 4t 6t a = f ''(t) 48t 6 7

8 Note that the first erivative of the velocity function is also the secon erivative of the position function. Using erivative notation, position: 8t t 1 velocity: v 4t 6t y v acceleration: a 48t 6 t t Alternate notation for the secon erivative: f ''(t) t notation for the secon erivative. oes not inicate a ratio of squares, but only the 8

9 Table of Derivatives Most are Provie on AP Physics Equation Sheet 1. ( const.) 0. ( 1. ( n ) n n1 4. cf ( c f ( 5. f ( g( f ( g( 6. f ( g( f ( g( (The Prouct Rule) 7. f ( g( f ( g( g( f ( 8. f ( g( g( f ( f ( g( g( (The Quotient Rule) 9. (sin cos 10. (cos sin 11. (tan sec (ln 1 e e 9

10 Derivative Practice y For problems 1-8 fin the erivative of y with respect to (i.e. ). Complete the assignment on a separate piece of graph paper. 1. y 5. 4 y. 8 4 y y ( 4 )( ) 5. y y 5 7. y sin cos 8. y 5cos 9. For the equation y 4 fin (a) the equation of the slope of its tangent line at any point. (b) the equation of the tangent line at point (4,) using point-slope form. 10. A particle unergoes straight-line motion with its isplacement at any time given by the following equation (assume position in meters, an time in secons) y t 4t t 1 (a) Fin the times when the particle is motionless. (b) Fin the time when the particle is moving to the right. (c) Fin the time when the particle is moving to the left. 11. The velocity of a particle moving along the -ais for t 0 is given by v 4 t. (a) What is the particle s acceleration when it first achieves a velocity of zero? (b) What is the particle s acceleration when it achieves its maimum isplacement in the +-irection? 1. The position of a particle moving along the -ais is given by 6t t 4. (a) What is the particle s velocity at times, t an t 4? (b) What is the particle s average acceleration from t to t 4 10

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