Quiz 7 (Solution)

Date: MATH 1210-4 - Spring 2004

Note: Unless stated otherwise, answers without justification receive no credit. Please, show your work!

1. Use differentials to approximate $\sqrt{62}$.

   Since $62$ is close to $64$, whose square root is $8$, we use the formula

   $$y(x_0 + \Delta x) \simeq y(x_0)+ y'(x_0) \Delta x,$$

   
   
   for $y(x)=\sqrt{x}$, $x_0=64$, and $\Delta x = -2$. Then, since $y'(x) = \frac{1}{2\sqrt{x}} \ensuremath{\Rightarrow}\xspace y'(64) = \frac{1}{2\sqrt{64}} = \frac{1}{16}$ we have

   $$\sqrt{62} = y(64-2) \simeq y(64) + y'(64) (-2) = 8 - \frac{2}{16} = \frac{63}{8} = 7.875.$$

   
   

   Using the calculator we obtain the value (rounded to five decimal places) $7.87401$.

2. An airplane, flying horizontally, at an altitude of 1 mile, passes directly over an observer. If the constant speed of the airplane is 600 miles per hour, how fast is its distance from the observer increasing 1 minute later?

   Figure 1: Schematic description for the airplane problem

   

   Figure 1 shows the graph where $s$ is the distance from the airplane to the observer and $x$ is the (horizontal) distance traveled by the airplane from the moment it passed over the observer.

   We know that $\frac{dx}{dt} = 600$ and we want to know $\frac{ds}{dt}$ one minute after the plane flew over the observer. Since the plane travels $\frac{600}{60} = 10$ miles per minute, we want to know $\frac{ds}{dt}$ when $x=10$.

   For all times we have the relation $s^2 = 1^2 + x^2$, so that, taking derivatives (with respect to time, $t$) on both sides we get

   $$2s\frac{ds}{dt} = 2x\frac{dx}{dt},$$

   
   
   so that

   $$\frac{ds}{dt} = \frac{x}{s} \frac{dx}{dt}.$$

   
   
   Now we see that when $x=10$, $s=\sqrt{1^2+10^2} = \sqrt{101} \simeq 10.049876$, and we obtain

   $$\frac{ds}{dt} = \frac{x}{s} \frac{dx}{dt} = \frac{10}{10.049876} \cdot 600 = 597.02231.$$