Answer
a) seconds/meter
b) L = 3 m
c) The instantaneous rate of change at L = 3 m is approximately 0.4330 s/m
Work Step by Step
a)
$\frac{ΔT}{ΔL}$ = $\frac{seconds}{meter}$ = seconds/meter
b) L = 3 m
c)
time interval [3, 3.01]
$\frac{ΔT}{ΔL}$ = $\frac{\frac{3\sqrt {3.01}}{2}-\frac{3\sqrt {3}}{2}}{3.01-3}$ = $0.4327$
time interval [3, 3.001]
$\frac{ΔT}{ΔL}$ = $\frac{\frac{3\sqrt {3.001}}{2}-\frac{3\sqrt {3}}{2}}{3.001-3}$ = $0.4330$
time interval [3, 3.0001]
$\frac{ΔT}{ΔL}$ = $\frac{\frac{3\sqrt {3.0001}}{2}-\frac{3\sqrt {3}}{2}}{3.0001-3}$ = $0.4330$
time interval [2.99, 3]
$\frac{ΔT}{ΔL}$ = $\frac{\frac{3\sqrt {3}}{2}-\frac{3\sqrt {2.99}}{2}}{3-2.99}$ = $0.4334$
time interval [2.999, 3]
$\frac{ΔT}{ΔL}$ = $\frac{\frac{3\sqrt {3}}{2}-\frac{3\sqrt {2.999}}{2}}{3-2.999}$ = $0.4330$
time interval [2.9999, 3]
$\frac{ΔT}{ΔL}$ = $\frac{\frac{3\sqrt {3}}{2}-\frac{3\sqrt {2.9999}}{2}}{3-2.9999}$ = $0.4330$
The instantaneous rate of change at L = 3 m is approximately 0.4330 s/m
