Answer
The instantaneous rate of change is approximately 0.57735 m/(s · K)
Work Step by Step
time interval [300,300.01]
$\frac{∆v}{∆T}$ = $\frac{v(300.01)-v(300)}{300.01-300}$ = $\frac{20\sqrt {300.01}-20\sqrt {300}}{300.01-300}$ = $0.577345$
time interval [300,300.005]
$\frac{∆v}{∆T}$ = $\frac{v(300.005)-v(300)}{300.005-300}$ = $\frac{20\sqrt {300.005}-20\sqrt {300}}{300.005-300}$ = $0.577348$
time interval [300,300.001]
$\frac{∆v}{∆T}$ = $\frac{v(300.001)-v(300)}{300.001-300}$ = $\frac{20\sqrt {300.001}-20\sqrt {300}}{300.001-300}$ = $0.57735$
time interval [300,300.00001]
$\frac{∆v}{∆T}$ = $\frac{v(300.00001)-v(300)}{300.00001-300}$ = $\frac{20\sqrt {300.00001}-20\sqrt {300}}{300.00001-300}$ = $0.57735$
The instantaneous rate of change is approximately 0.57735 m/(s · K)
