• If the car travels for $2 \, hours$, then it moves $5 \, miles + 5 \, miles = 10 \, miles$, or $\frac {5 \, miles}{1 \, hour} \cdot 2 \, hours = 10 \, miles$
• If the car travels for $5 \, hours$, then it moves $\frac {5 \, miles}{1 \, hour} \cdot 5 \, hours = 25 \, miles$
• If the car travels for $\frac {1}{2} \, hour$, then it moves $\frac {5 \, miles}{1 \, hour} \cdot \frac {1}{2} \, hours = 2.5 \,miles$

The way we calculate distance is by mulitplying the rate times the length of time.

We can model this graphically.

Area = Accumulated Distance

$\blacktriangleright$ A Constant Rate

Our car’s rate is a constant function: $r(t) = 5$.

Since time is measured by the horizontal axis. our $distance = rate \times time$ calculation can be represented visually as the rectangular area under the rate curve.

The area of the rectangle represents the distance travelled.

The height of the rectangle is the height of the graph, which is the value of the rate, $5 \, mph$.

The width of the rectangle is a time measurement. In the graph above, this is $1 \, hour$.

The area is height times width, which gives $area = 5 \, mph \times 1 \, hour = 5 \, miles$

Let $D(t) =$ the accumulated miles travelled after $t \, hours$.

Graphically, this is the area of the rectangle below the graph from $0$ to $t$.

$D(t)$ is the accumulated distance travelled in $t \, hours$, which makes $r(t)$ its derivative.

$$D'(t) = r(t)$$

$$(5 t)' = 5$$

$\blacktriangleright$ A Linear Rate

This time our rate function is a linear function: $r(t) = 2 t$.

Our car goes faster as it goes farther.

$Distance$ is still represented visually as the area under the rate curve.

The area is a triangle this time.

Over the interval $[0,t]$, the area is $D(t) = \frac {1}{2} \cdot t \cdot (2 t) = t^2$

$D(t) = t^2$ is the accumulated distance travelled in $t \, hours$, which makes $2 t$ its derivative.

$$(t^2)' = 2 t$$

Accumulation

Below is the graph of $y = p(k) = -k^2 + 7$, approximated with

$$L(k) = \begin {cases} 4(k+2)+3 & \text { on } \, \left (-3, -\frac {3}{2}\right ] \\ 2(k+1)+6 & \text { on } \, \left (-\frac {3}{2}, -\frac {1}{2}\right ] \\ 7 & \text { on } \, \left (-\frac {1}{2}, \frac {1}{2}\right ] \\ -2(k-1)+6 & \text { on } \, \left (\frac {1}{2}, \frac {3}{2}\right ] \\ -4(k-2)+3 & \text { on } \, \left (\frac {3}{2}, 3\right ) \end {cases}$$

$L(k)$ is a piecewise linear function, which means we can calculate area using rectangles and triangles.

The shaded regions are made of triangles, trapezoids, and rectangles. And, we know their formulas from Geometry class.

$$\text {Area} = \frac {1}{2} \cdot 1.25 \cdot L(-1.5) + \frac {1}{2} \cdot (L(-1.5) + L(-0.5)) \cdot 1 + 1 \cdot L(0.5) + \frac {1}{2} \cdot (L(1.5) + L(0.5)) \cdot 1 + \frac {1}{2} \cdot 1.25 \cdot L(1.5)$$

$$\text {Area} = \frac {1}{2} \cdot 1.25 \cdot 5 + \frac {1}{2} \cdot 12 \cdot 1 + 1 \cdot 7 + \frac {1}{2} \cdot 12 \cdot 1 + \frac {1}{2} \cdot 1.25 \cdot 5$$

$$\text {Area} = 25.25$$

This is the area under $y = L(x)$, so it is the approximate area under $y = p(k) = -k^2 + 7$ from $k = -3$ to $k = 3$.