## The Second Derivative and the Shape of a Function f(x)

Read this section to learn how the second derivative is used to determine the shape of functions. Work through practice problems 1-9.

### Feeling the Second Derivative

Earlier we saw that if a function $\mathrm{f}(\mathrm{t})$ represents the position of a car at time $\mathrm{t}$, then $\mathrm{f}^{\prime}(\mathrm{t})$ is the velocity and $f'' (\mathrm{t})$ is the acceleration of the car at the instant $\mathrm{t}$.

If we are driving along a straight, smooth road, then what we feel is the acceleration of the car:

a large positive acceleration feels like a "push" toward the back of the car,
a large negative acceleration (a deceleration) feels like a "push" toward the front of the car, and an acceleration of $0$ for a period of time means the velocity is constant and we do not feel pushed in either direction.

On a moving vehicle it is possible to measure these "pushes", the acceleration, and from that information to đetermine the velocity of the vehicle, and from the velocity information to determine the position. Inertial guidance systems in airplanes use this tactic: they measure front-back, left-right and up-down acceleration several times a second and then calculate the position of the plane. They also use computers to keep track of time and the rotation of the earth under the plane. After all, in 6 hours the earth has made a quarter of a revolution, and Dallas has rotated more than 5000 miles!

Example 1: The upward acceleration of a rocket was $\mathrm{a}(\mathrm{t})=30 \mathrm{~m} / \mathrm{s}^{2}$ for the first 6 seconds of flight, $0 \leq \mathrm{t} \leq 6$. The velocity of the rocket at $\mathrm{t}=0$ was $0 \mathrm{~m} / \mathrm{s}$ and the height of the rocket above the ground at $\mathrm{t}=0$ was $25 \mathrm{~m}$. Find a formula for the height of the rocket at time $\mathrm{t}$ and determine the height at $\mathrm{t}=6$ seconds.

Solution: $\mathrm{v}^{\prime}(\mathrm{t})=\mathrm{a}(\mathrm{t})=30$ so $\mathrm{v}(\mathrm{t})=30 \mathrm{t}+\mathrm{K}$ for some constant $\mathrm{K}$. We also know $\mathrm{v}(\mathbf{0})=0$ so $30(0)+K=0$ and $K=0$. Therefore, $v(t)=30 t$
Similarly, $h^{\prime}(t)=v(t)=30 t$ so $h(t)=15 t^{2}+C$ for some constant $C$. We know that $h(0)=25$ so $15(0)^{2}+C=25$ and $\mathrm{C}=25$. Then $\mathrm{h}(\mathrm{t})=15 \mathrm{t}^{2}+25$. $\mathrm{h}(6)=15(6)^{2}+25=565 \mathrm{~m}$