Unit 4 practice test: Contextual Applications of Differentiation

The slope of the tangent line is 3; the secant from $(1, 1.5)$ to $(3, 5.5)$ has slope 2. The population grows faster at $t = 3$ than on average before it.

• $P'(3)$ equals the average rate, since both use $t = 3$: The average rate is the slope of a secant line through two points; it is 2, not 3.
• $P'(3)$ is less than the average rate, which is $4$: That divides $5.5 - 1.5 = 4$ by 1 instead of by 2 hours.
• Correct: $P'(3)$ is greater: 3 versus an average of 2: The average rate is $\frac{5.5 - 1.5}{3 - 1} = 2$.
• They are not comparable, because their units differ: Both are in thousand bacteria per hour.

$P'(3) = 3$ means that at $t = 3$ hours the number of bacteria is increasing at 3 thousand bacteria per hour.

• After 3 hours, there are 3 thousand bacteria: That would mean $P(3) = 3$; the graph shows $P(3) = 5.5$.
• Each hour, the population triples: The derivative measures a rate of change in thousands per hour, not a growth factor.
• During the first 3 hours, a total of 3 thousand bacteria were added to the culture: That is a total change over an interval; $P'(3)$ is a rate at one instant.
• Correct: At 3 hours, the population is growing at 3 thousand bacteria per hour: Instant, quantity, direction and units are all named.

Values of $P$ are counts (thousands); slopes, derivatives and difference quotients are rates (thousands per hour).

• Correct: $P(3)$: A value of $P$ is a number of bacteria, in thousands.
• $P'(3)$: A derivative of $P$ is a rate, in thousands per hour.
• The slope of the tangent line at $t = 3$: A slope on this graph is thousands per hour.
• $\frac{P(5) - P(3)}{5 - 3}$: A change in $P$ divided by a time is a rate.

As $x \to 0$, $1 - \cos x \to 0$ and $x^2 \to 0$: the indeterminate form $\frac{0}{0}$.

• $\frac{1}{0}$: $\cos 0 = 1$, so the top is $1 - 1 = 0$.
• $\frac{\infty}{\infty}$: Neither part grows without bound near 0.
• $\frac{2}{0}$: $1 - \cos 0 = 0$, not 2.
• Correct: $\frac{0}{0}$: $1 - \cos 0 = 0$ and $0^2 = 0$.

$\sin 0 = 0$ and $2(0) = 0$: still $\frac{0}{0}$. Apply the rule again (both limits are 0): $\lim_{x \to 0}\frac{\cos x}{2}$$= \frac{1}{2}$.

• Correct: Still $\frac{0}{0}$, so apply the rule again to get $\frac{1}{2}$: Check the form at every round; a second application finishes it.
• It equals 0, because $\sin 0 = 0$: The bottom is also 0, so the form is still indeterminate.
• It does not exist, because the bottom is 0 at $x = 0$: A zero bottom with a zero top is indeterminate, not a missing limit.
• Use the quotient rule on $\frac{\sin x}{2x}$, then substitute 0: L'Hospital's rule differentiates top and bottom separately; the quotient rule is not part of it.

For a $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, $\lim\frac{f(x)}{g(x)}$$= \lim\frac{f'(x)}{g'(x)}$, when the second limit exists.

• $\lim \left(\frac{f(x)}{g(x)}\right)'$: That is the derivative of the quotient, a different function.
• Correct: $\lim \frac{f'(x)}{g'(x)}$: Differentiate the top and the bottom separately.
• $\lim \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$: That is the quotient rule, which is not part of L'Hospital's rule.
• $\lim \frac{g'(x)}{f'(x)}$: The derivatives keep their places: $f'$ on top.