1. ## 2 Calculus Questions!!!

1. At noon, ship A is 20 nautical miles due west of ship B. Ship A is sailing west at 23 knots and ship B is sailing north at 22 knots. How fast (in knots) is the distance between the ships changing at 6 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Below is the graph of the derivative of a function defined on the interval (0,8). You can click on the graph to see a larger version in a separate window.

2.

Refer to the graph to answer each of the following questions. For parts (A) and (B), use interval notation to report your answer. (If needed, you use U for the union symbol, and write None if there is no answer.)

(A) For what values of in (0,8) is increasing? (0,4)U(5,4) ?
(B) For what values of in (0,8) is concave down? (4,3) ?
(C) Find all values of in (0,8) is where has a local minimum? None ?
(D) Find all values of in (0,8) is where has an inflection point?? 4 ?

Help me please see if i did it right!! thanks.

2. For number 1 ..is this the right way to do it??/

d(t)^2 = (23t + 20)^2 + 22t^2 (pythag)

d'(t)^2 = 2* 23^2*t + 2*22^2*t

Let t = 6

2*23^2*6 + 2*22^2*6 = 12 ( 23^2 + 22^2) = 12156

d'(6)^2 = 12156
d'(6) = 110.3kph

(Note: 1 knot is a speed of 1 nautical mile per hour.) ???

3. You are very close. You start out good:

$\displaystyle d(t)^2 = (20+23t)^2+(22t)^2$

But you go wrong on the next line, it should be:

$\displaystyle 2d(t)d'(t)=2(20+23t)(23)+2(22t)(22)$
$\displaystyle d(t)d'(t)=(20+23t)(23)+(22t)(22)$
$\displaystyle d'(t)=\frac{(20+23t)(23)+(22t)(22)}{d(t)}$

Then,

$\displaystyle d'(t)=\frac{(20+23t)(23)+(22t)(22)}{\sqrt[]{(20+23t)^2+(22t)^2}}$

Then,

d'(6) = $\displaystyle \frac{3269}{\sqrt{10597}} \approx 31.755$ knots

4. Thanks so much Memtia..i really appreciated.