Rate of Change and Tangents

Rate of Change & Tangents

We'll consider the most fundamental concepts in calculus: limits. We will also look at continuity. Continuity is the most fundamental concept in calculus. Now a physical example.

Standing at the edge of a cliff, I release a rock. As it falls over the edge and begins to fall down the cliff, its speed increases until it hits the bottom. We know that this piece of rock will fall faster and faster until it hits the bottom at which point its speed is constant.

The question is: how can we find out exactly how fast this rock is traveling at any given moment during its descent?

From physics, we know that the displacement of a released object (the piece of rock) after t seconds is given by the formula y=4.9t^2

y=4.9t2

Probably you already recognize why 4.9g is equal to 1/2 times the gravitational constant, which is 9.8 m/s2. We have 4.9 times t squared.

All right, let's consider the instantaneous velocity at time t.

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Instead of performing that calculation, let's consider a simple situation: After two seconds have elapsed, we know that the rock's position will be represented by this point...

t=2 sec

The displacement after two seconds is given by 4.9 times t squared, where t equals two. So it is 19.6 metres.

y=4.9(22)=19.6m

Thus, it is obvious that the average velocity of the piece of rock from t equals zero to t equals two is 9.8 metres per second.

Now, we can do the same thing with the velocity of a piece of rock moving in another time frame, namely between t = 1 and t = 2.

Now, in this situation, we must first compute the distance the rock has fallen through in this period of time.

Therefore, the displacement at time t equals to 4.9 times two square metre minus the displacement at time t equals to one, which is 4.9 times one square metre.

The distance the rock has travelled in this time frame is equal to two minus one, and you divide by the time it spans in travelling this distance, namely two minus one.

This gives us the average velocity of a rock from time t equals one to time t equals two. We can use this equation for any time frame by substituting in the appropriate values.

Now imagine we start with the time point t seconds.

At time t seconds, the rock is at a height of 4.9t square meters from the top. If h seconds have elapsed, then the next time point is t + h seconds, and the displacement from the beginning is 4.9t + h square meters.

As in the previous example, we can now compute average velocity of the piece of rock over this time frame. The average velocity is equal to the displacement over the time spanned during this period, t plus h minus t.

The average velocity over the time interval from t-zero to t-zero plus h is given by this fraction.

Displacement at time t =129. 8t2( ) = 4. 9t2

Average velocity over the time interval t0,t0[ + h] is given by

4.9(t0+h)2−4.9t02h =4.9 t02+2t0h+h2( )−4.9t

To find the instantaneous velocity at t = 0, we can use the average velocity to approximate the instantaneous velocity by considering h smaller and smaller.

t0=1

In this table, we see the values of h, starting at one and gradually decreasing to .5, .1 second, .01 second and so on. On the right-hand side of the table are corresponding values for average velocity: 14.7 meters/second, 12.25 meters/second and so on.

We now consider the average velocity of a particle at time t--the average velocity, that is, over an interval of length h--and we hope to bring it closer and closer to the instantaneous velocity at t=0 by reducing h. We can do this as long as we don't put h equal to zero because if we did, then we'd get 0/0 which is meaningless.

However, remember that we are interested in what happens as h gets smaller and smaller, so this fraction can be simplified into this form.

To find the instantaneous velocity at time t = 0, we multiply out the things in the numerator, cancelling 4.9 t-zero square, and then divide by h. We get this expression. As h goes very close to zero, this value goes very close to 9.8 t-zero:

So we've seen the example of trying to find the instantaneous velocity of a piece of rock that falls through a cliff. Now, if we go back to the process we used in doing that, we see that we start with finding an average velocity over a small time window, then try to approach the instantaneous velocity by reducing this window's width--so using this same approach, we can now do more generally for a function y=f(x).

y=f(x)

In the previous example, y is a function of time t, which is the displacement of the piece of rock. So generally, we can look at other situations in which y depends on x in some way, which is written now as a function of x. Following our previous example, we can try to find

the average change of the function y over the change of the variable x by dividing delta y by delta x.

This formula represents the average rate of change of y with respect to x, so as in the previous example, we compute in the numerator the change in f from xA to xB, and then divide by the change in x, namely xB minus xA.

Thus, we can write this rate of change in the following form: if h denotes xB - xA, then h must be non-zero.

In order to understand the meaning of this average change of the function, we can try to interpret graphically what it represents.

Here is a graph.

The blue line represents the graph of the function y equals f(x). This point here is the point with x-coordinate xA and corresponding y-coordinate f(xA). The height of this point is then xB and its neighbouring point is represented by f(xB).

Thus, the average change of the function equals to the vertical distance between f(xB) and f(xA) divided by the horizontal distance between xB and xA. That means h.

If you divide the vertical height by the horizontal distance, you get the slope of this straight line--we call it the secant--joining these two points, xA and xB.

We have the slope of the secant through point A and B is equal to this ratio, f(xB) - f(xA) divided by xB - xA, and also equal to this: h = xB - xA.

Seems like this is the average rate of change of f(x) over this interval xA-xB,

Let us consider what happens when point xB approaches xA. This means that we are trying to reduce the value of h so that point xB moves toward xA. You can see clearly that

the line joining the two points becomes this one and this one, so gradually, you can see very clearly that this line would become the tangent line at point xA.

This definition is called the derivative of the function y at the point x-one, so we have this very natural relationship. The instantaneous rate of change of y, which we call its derivative, equals the slope of the tangent line at x-one.