MAT1193 – 4c Definition of derivative With a better understanding of limits we return to idea of the instantaneous velocity or instantaneous rate of change. Remember that in the example of calculating the speed that I was going on my bike, what we used the formula average rate =
Δposition
Δtime
to calculate the average rate of change, and chose smaller and smaller time intervals Δt to get better and better approximations. €
Let’s consider a more specific example. Suppose a young kid throws a ball up in the air. Let h be the height of the ball in meters and t be the time after the ball has been thrown in seconds. The height is a function that depends on time is give by the formula h(t) = -­‐5t2+10t Suppose we want to know how fast the ball is traveling exactly 1.5 seconds after it is thrown. To start a decent approximation of that number would be to find the average velocity from 1.5 to 1.6 seconds. Plugging in those values of time we find h(1.5) = 3.75 m and h(1.6) = 3.2 average rate =
€
So the average rate of change over that 0.1 seconds is -­‐5.5 meters per second. Note that a negative velocity means that the ball is already on its way down. We can get a better approximation by considering the velocity from t=1.5 to t = 1.51 seconds. Since h(1.51) = 3.6995, we have average rate =
€
Δh 3.6995 − 3.75 −.0505
=
=
= −5.05 Δt
1.51 −1.5
.01
An even better approximation can be had by considering the average velocity for just 1/1000th of a second, from t=1.5 to t=1.501. With h(1.501)= 3.7450, the average velocity is average rate =
€
Δposition Δh 3.2 − 3.75 −.55
=
=
=
= −5.5 Δtime
Δt
1.6 −1.5
.1
Δh 3.7450 − 3.75 −.0050
=
=
= −5.00 Δt
1.501 −1.5
.001
If we kept going with this we’d find number getting closer and closer to -­‐5 m/s. (I only took the last answer out to 2 decimal places. If I did more, the answer would be really close to, but not equal to, -­‐5 m/s.) In other words, we have lim
Δt → 0
Δh
= −5 Δt
In the limit of really small intervals the change in height and the change in time get infinitesimally small. When that happens we use a lower case d instead of the Δ and €
dh
= −5 . This is the derivative of the function h with respect to t. One thing dt
that is missing from this notation is the fact that the entire calculation focused on finding the velocity when t=1.5. If we found the velocity at a different time, we’d obviously get a different answer. We call 1.5 seconds the base point for our € calculation, and since it represents a specific time we usually write t =1.5. Then, to 0
specify that we are calculating the derivative at time t0 we write write dh
= −5 dt t 0 =1.5
€
This is called the differential notation for the derivative. Writing the derivative with this notation emphasizes the roots of the derivative as a limit of average rates of change or equivalently as a limit of slopes of secant lines. We can also treat the derivative as just another function: you give me a base point, I calculate the derivative and spit out the answer. If we want to treat the derivative that way we generally use the prime notation h’(1.5) = -­‐5. How do we think about the derivative in terms of a graph? Let’s plot the function h(t) = -­‐5t2+10t (the slightly curved blue line). 4.2
4
6t
h(t)
3.8
3.6
6h
3.4
3.2
3
1.4
1.45
1.5
1.55
t
1.6
1.65
Lets look at our first approximation, using t=1.5 and t=1.6 and find what in the graph corresponds to the average rate of change? According to the definition, average rate =
€
Δh
Δt
This is just the slope of the secant line through the points (1.5, 3.75) and (1.6, 3.2) (the straight black line). A secant line for a given function is just a line that goes through two points on the graph of that function. (Remember that the point (1.5, 3.75) is obtained by plugging t=1.5 into the function h(t) = -­‐5t2+10t, and the point (1.6, 3.2) is obtained by plugging t=1.6 into h(t).) What happens to the graph as we try to get a better approximation to the instantaneous rate of change? Like we did before, let’s keep the same base point (1.5, 3.75) but reduce Δt to .01. Then the picture looks like Now the secant line passes through the base point (1.5, 3.75) marked by the dot and the second point (1.51, 3.6995) marked by the open circle. As before the average rate of change which is just the slope of this line and is equal to average rate =
€
Δh 3.6995 − 3.75 −.0505
=
=
= −5.05 Δt
1.51 −1.5
.01
What is the limit of this process as we get better and better approximations? As Δt gets infinitely close to 0, then the open circle approaches the base point and the secant lines that we’ve been calculating approach a special line called the tangent line. 4.2
4
h(t)
3.8
3.6
3.4
3.2
3
1.4
1.45
1.5
1.55
1.6
1.65
t
The tangent line is defined as the line going through the base point and whose slope is the limit of the slopes lim
Δt → 0
Δh
Δt
dh
is equal to the slope of dt t 0 =1.5
the tangent line passing through the base point (1.5,h(1.5)). In other words, the derivative, written as h’(1.5) or €
Why is calculating the derivative and finding the tangent line such a big deal? The € for a line is really simple and (b) as most important reason is that the (a) formula long as you stay pretty close to the base point, this really simple function gives output values that are really close to our original (and potentially very complicated!) function. In terms of our problem this means the following. We know that after 1.5 seconds the ball thrown by the young kid is 3.75 meters in the air. We just figured out that at that time it’s traveling at -­‐5 meters per second. So what’s the height at 1.49 seconds? Well we could plug 1.49 into the original formula and find h(1.49) = -­‐5*(1.49)2+10*1.49= 3.7995. Alternatively we can simplify the problem by assuming that the ball is traveling at approximately -­‐5 meters per second for times near 1.5 seconds. Near that time, the derivative is a good approximation for the average rate of change: dh Δh
≈
dt Δt
where the wavy lines mean “approximately equal to.” Then since €
Δh =
€
Δh
* Δt we can write Δt
Δh ≈
dh
* Δt
dt
Now, since 1.49 seconds is .01 seconds before 1.5, so we can think of 1.5 as the base €
point t0, and the difference between 1.49 and 1.5 = -­‐0.01 as the change in time Δt. Since the derivative at t0 is equal to -­‐5, we find that the change in height is approximately (-­‐5)*(-­‐.01) = 0.05 meters. Since the height at the base point was 3.75, we approximate that the height of the ball at time 1.49 should be 3.75+0.05= 3.8 meters. This is really close to the actual height of 3.7995 and the calculation involved was much simpler. Later, we’ll see that the derivative is not only useful for simplifying calculations, but is also very useful for characterizing the relative rates of change between variables and how that rate of changes itself changes as we move the base point.