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Average rate of change on $[1,e]$

Let h(t)=ln⁡(t)h(t) = \ln(t).

Consider h(t)=ln⁡(t)h(t) = \ln(t). Find the average rate of change of hh on the interval [1,e][1, e].

Hint

The average rate of change on [a,b][a,b] is h(b)−h(a)b−a\dfrac{h(b) - h(a)}{b - a}. Recall that ln⁡(1)=0\ln(1) = 0 and ln⁡(e)=1\ln(e) = 1.

Step-by-step solution

The average rate of change of hh on [1,e][1,e] is given by the slope of the secant line through (1,h(1))(1, h(1)) and (e,h(e))(e, h(e)): h(e)−h(1)e−1.\frac{h(e) - h(1)}{e - 1}.

Since h(1)=ln⁡(1)=0h(1) = \ln(1) = 0 and h(e)=ln⁡(e)=1h(e) = \ln(e) = 1, this becomes 1−0e−1=1e−1.\frac{1 - 0}{e - 1} = \frac{1}{e-1}. So the average rate of change is 1e−1\frac{1}{e-1}.

Instantaneous rate of change at $t=e$

Use the limit definition of the derivative, h′(a)=lim⁡t→ah(t)−h(a)t−a,h'(a) = \lim_{t \to a} \frac{h(t) - h(a)}{t - a}, to find the instantaneous rate of change of h(t)=ln⁡(t)h(t) = \ln(t) at t=et = e.

Hint

Set up the limit lim⁡t→eln⁡(t)−ln⁡(e)t−e=lim⁡t→eln⁡(t)−1t−e\lim_{t \to e} \dfrac{\ln(t) - \ln(e)}{t - e} = \lim_{t \to e} \dfrac{\ln(t) - 1}{t-e}, and use the substitution t=e1+ut = e^{1+u} (or a table of values) to see what value this approaches as t→et \to e.

Step-by-step solution

By the limit definition, h′(e)=lim⁡t→eln⁡(t)−ln⁡(e)t−e=lim⁡t→eln⁡(t)−1t−e.h'(e) = \lim_{t \to e} \frac{\ln(t) - \ln(e)}{t - e} = \lim_{t \to e} \frac{\ln(t) - 1}{t - e}. Writing t=e+ht = e + h so that t→et \to e corresponds to h→0h \to 0, this is lim⁡h→0ln⁡(e+h)−1h.\lim_{h\to 0} \frac{\ln(e+h) - 1}{h}.

Evaluating this limit numerically (using values of tt close to e≈2.71828e \approx 2.71828 from both sides) shows the difference quotient approaches 1e≈0.3679\frac{1}{e} \approx 0.3679. Thus the instantaneous rate of change of hh at t=et=e is h′(e)=1eh'(e) = \frac{1}{e}.

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