  ### Home > CALC > Chapter 5 > Lesson 5.2.1 > Problem5-57

5-57.

After a weather balloon is launched the following data is sent back to the meteorological center. The column marked time, $t$, is given in seconds after launch, elevation, $e$, in feet above sea level, and temperature, $T$, in degrees Fahrenheit.

1. Approximately how fast (in ft/sec) is the balloon rising at $120$ seconds?

Since we cannot find the derivative (IROC), calculate the average rate of change.
For $e(t)$:
You could use Hana’s method: $\frac{e(150)-e(120)}{150-120}$
or Anah’s method: $\frac{e(120)-e(90)}{120-90}$
or Hanah’s method: $\frac{e(150)-e(90)}{150-90}$

2. Approximately how fast is the temperature changing (in °F/ft) at $2750$ feet?

Use Hana, Anah, or Hanah's method to approximate the rate of change for $T(e)$ at $e = 2750$.

3. Approximately how fast is the temperature changing (in °F/sec) at $120$ seconds?

Use Hana, Anah, or Hanah's method to approximate the rate of change for $T(t)$ at $t = 120$.

4. Why would you expect the product of the answers to (a) and (b) to equal answer (c)?

Rewrite: $e'(t)=\frac{de}{dt} \text{ and }T'(e)=\frac{dT}{de}$
$\frac{de}{dt}\cdot \frac{dT}{de}=$

$t$ (Time)

$e$ (Elevation)

$T$ (Temp)

$0$

$1260$

$56.4°$

$30$

$1560$

$55.1°$

$60$

$1920$

$53.9°$

$90$

$2350$

$51.9°$

$120$

$2750$

$49.5°$

$150$

$3170$

$47.7°$

$180$

$3600$

$45.0°$

$210$

$4080$

$42.4°$

$240$

$4560$

$40.9°$