Q: If $$N(t) = \frac{180t}{t + 4}$$, what is the approximate number of subscribers (in thousands) after 16 months?

$$N(16) = \frac{180 \times 16}{16 + 4} = 144$$

Q: Using $$N'(t) = \frac{180}{t + 4} - \frac{180t}{(t + 4)^2$}$$, what is the rate of change of subscribers at $$t = 16$$ months?

$$N'(16) = \frac{180}{20} - \frac{180 \times 16}{400} = 9 - 7.2 = 1.8$$

Q: What is the rate of change of drug concentration in the left arm at $$t = 1$$ hour, given $$C'(t) = -\frac{0.28t}{(t^2$ + 1)^2$}$$?

$$C'(1) = -\frac{0.28 \times 1}{(1^2$ + 1)^2$} = -\frac{0.28}{4} = -0.07$$

Q: For $$C(t) = \frac{0.14}{t^2$ + 1}$$, what is the rate of change of concentration at $$t = 3$$ hours?

$$C'(3) = -\frac{0.28 \times 3}{(3^2$ + 1)^2$} = -\frac{0.84}{100} = -0.0084$$

Question 1

A communications company has installed a new cable television system in a city. The total number $$N(t) (in $$thousands) of subscribers $$t$$ months after installation is given by $$N(t) = \frac{180t}{t + 4}. $$What does $$N'(t)$$ represent in this context?

Accepted Answer

The rate at which the number of subscribers is increasing per month.

Question 2

Given $$N(t) = \frac{180t}{t + 4}$$, what is $$N'(t)$$?

Accepted Answer

$$N'(t) = \frac{180}{t + 4} - \frac{180t}{(t + 4)^2$}$$

Question 3

If $$N(t) = \frac{180t}{t + 4}$$, what is the approximate number of subscribers (in thousands) after 16 months?

Accepted Answer

$$N(16) = \frac{180 	imes 16}{16 + 4} = 144$$

Question 4

Using $$N'(t) = \frac{180}{t + 4} - \frac{180t}{(t + 4)^2$}$$, what is the rate of change of subscribers at $$t = 16$$ months?

Accepted Answer

$$N'(16) = \frac{180}{20} - \frac{180 	imes 16}{400} = 9 - 7.2 = 1.8$$

Question 5

A drug is injected into a patient’s bloodstream. The concentration in the left arm is given by $$C(t) = \frac{0.14}{t^2$ + 1}. $$What does $$C'(t)$$ represent?

Accepted Answer

The rate at which the drug concentration is changing in the left arm over time.

Question 6

Find $$C'(t)$$ for $$C(t) = \frac{0.14}{t^2$ + 1}.$$

Accepted Answer

$$C'(t) = -\frac{0.28t}{(t^2$ + 1)^2$}$$

Question 7

What is the rate of change of drug concentration in the left arm at $$t = 1$$ hour, given $$C'(t) = -\frac{0.28t}{(t^2$ + 1)^2$}$$?

Accepted Answer

$$C'(1) = -\frac{0.28 	imes 1}{(1^2$ + 1)^2$} = -\frac{0.28}{4} = -0.07$$

Question 8

If $$C(t) = \frac{0.14}{t^2$ + 1}$$, what is the drug concentration at $$t = 0.5$$ hours?

Accepted Answer

$$C(0.5) = \frac{0.14}{0.25 + 1} = \frac{0.14}{1.25} = 0.112$$

Question 9

For $$C(t) = \frac{0.14}{t^2$ + 1}$$, what is the rate of change of concentration at $$t = 3$$ hours?

Accepted Answer

$$C'(3) = -\frac{0.28 	imes 3}{(3^2$ + 1)^2$} = -\frac{0.84}{100} = -0.0084$$

Question 10

If the rate of change of subscribers at $$t = 16$$ months is $$N'(16) = 1.8 (in $$thousands per month), estimate the total number of subscribers after 17 months.

Accepted Answer

$$N(17) \approx N(16) + N'(16) = 144 + 1.8 = 145.8$$

Business Calculus Study Guide: Applications of Derivatives

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