Topic 2.1 Extension 03 The derivative dy/dt To follow the worksheet “The derivative dy/dt” 1) In the lesson "Instantaneous velocity v" you learned how to find the instantaneous speed of a particle given a graph of its height vs. the time. To find the instantaneous speed of the particle we took the limit of y/t as t approached zero. The limit was the number that the ratio (or rate) y/t neared as t as got smaller and smaller. We call dy/dt the derivative of the function y(t) with respect to time t.

FYI: The instantaneous speed v is defined lim  y dy v =  t  0  t = dt .

2) Consider the function y(t) = 20 – 5t2 which describes the height y of a particle at time t. Let t1 = t be the time at which we would like to have the instantaneous speed of the particle. Let t2 be the time we will vary by allowing t to approach zero. Then we have

t1 = t (given)

t2 = t + t (since t2 – t1 = t) y1 = y(t1) = y(t) = 20 – 5t2 (substitution) y2 = y(t2) = y(t + t) = 20 – 5(t + t)2 (substitution)

so that

a) = y/t b) = [y2 - y1]/[t2 - t1] c) = [y(t2) - y(t1)]/[t2 - t1]

d) = [y(t + t) - y(t)]/[t + t - t] e) = [y(t + t) - y(t)]/t f) = [20 – 5(t + t)2 – (20 – 5t2)]/t g) = [20 – 5(t + t)2 – 20 + 5t2]/t h) = [20 – 5(t2 + 2tt + t2) - 20 + 5t2]/t i) = [20 – 5t2 – 10tt – 5t2 - 20 + 5t2]/t j) = [–10tt – 5t2]/t k) = –10t – 5t

Each of the above steps a) through k) should be justified; they have been drawn out in painful detail. For example, a) and b) are just the definition of average speed, and the definition of the symbol delta. Now you write the precise justification for each of the remaining letters:

1 c)______Substitute y2 = y(t2) and y1 = y(t1). d)______Substitute t2 = t + t and t1 = t. e)______Cancel the t's in the denominator. 2 f)______Substitute t + t and t into y(t) = 20 – 5t . 2 g)______Distribute the subtraction to the 20 and the -5t . 2 h)______Do FOIL on (t + t) . i)______Distribute the -5 to the t2 + 2tt + t2. 2 2 j)______Cancel 20 and -20, 5t and -5t . k)______Cancel a t from numerator and denominator.

3) Note that we now have a representation for the average speed of a particle whose height is given by y(t) = 20 – 5t2 at the time t and for

the time interval t, namely = –10t – 5t. At first sight it might seem that we did an awful lot of work to obtain this result, but consider the following:

i. is really a function of t. Thus we can find for any time t. ii. We can choose the time interval t to be anything we wish. Thus the limitations of a calculator have been overcome.

4) For example, suppose we want the average speed of the particle between the time t = 1.49 s and the time t = 1.50 s. Then t = 1.49 s and t = 1.50 – 1.49 = 0.01 s so that = –10(1.49) – 5(0.01) = -14.95 m/s.

a) Find the average speed of the particle for the time interval 0.00  t  0.01 s. t = 0.00 and t = 0.01 - 0.00 = 0.01 so that

= –10t – 5t = -10(0.00) - 5(0.01) = -0.05 m/s

b) Find the average speed of the particle for the time interval 1.50  t  1.51 s. t = 1.50 and t = 1.51 - 1.50 = 0.01 so that

= –10t – 5t = -10(1.50) - 5(0.01) = -15.05 m/s c) How do all these answers compare with those of 5) of the previous lesson? They are all the same.

5) To get an instantaneous speed, we let t shrink to zero. Thus

lim  y lim lim v =  t  0  t =  t 0 = t  0(–10t – 5t) = -10t .

Thus we have found the instantaneous speed v of the particle at any time we want! Now using this formula find the following instantaneous speeds:

v at t = 0.15 s v at t = 0.95 s v at t = 1.25 s v at t = 1.95 s v ______= -10(0.15) v ______= -10(0.95) v______= -10(1.25) v ______= -10(1.95) = -1.5 m/s = -9.5 m/s = -12.5 m/s = -19.5 m/s

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