BC Calculus 1
Rate of Change
Below is a graph of displacement vs. time. (Of course this is an approximation, since
physically speaking, a displacement graph cannot have “corners.”) Recall that
displacement is not the same as distance, since displacement can be negative, while
distance is always positive. For example, if you traveled east for 30 miles, then west 40
miles, you would have traveled a distance of 70 miles. However, considering east to be
the direction of positive displacement, your total displacement would have been 10
miles.
On the graph below, draw the corresponding velocity graph. Note that like displacement,
velocity may be negative. If the units for s are miles, what is the total distance traveled?
The total displacement?
s
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t
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
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Notes:
BC 1-2
Spring 2012
BC Calculus 1
Rate of Change
Below, you will find the graph of a displacement function.
1. Begin with a particular t value, and find the corresponding point on the curve.
2. Now take your straightedge (or perhaps the edge of a piece of paper) and place it
tangent to the curve at this point.
3. Using the grid lines, estimate the slope of this tangent line – call it m. Now plot the
point (t , m ) on the grid.
4. Repeat several times, so that you have plotted a set of points for which the second
coordinates are the slopes of the tangent lines at the corresponding t values.
5. Finally, draw a curve through these points. This is an approximate graph of the
corresponding velocity curve, which is the derivative of the displacement function.

s
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
t
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







Notes:
BC 1-2
Spring 2012
BC Calculus 1
Rate of Change
Below is a velocity graph. Using a starting point of (0,3) , draw the corresponding
displacement graph. Note that whether the endpoints of the segments on the velocity
graph are open or closed does not affect the displacement graph.
When you have finished making the graph, give a piecewise definition for both the the
given velocity curve and the displacement graph you have drawn.
v
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
t










BC 1-2
Spring 2012