1. ## Calculus..

If someone would take the time to help me, that'd be greatly appreciated.
I don't get any of it.

2. Originally Posted by Fnus
If someone would take the time to help me, that'd be greatly appreciated.
I don't get any of it.
number 13:
a)
AP=x meaning RP=2+x
using similar triangles, $\frac{AP}{AB} = \frac{RP}{QR}$

implying $\frac{x}{1} = \frac{2+x}{QR}$, so

$QR = \frac{2+x}{x}$

b) L(x) is the length..
using Pythagorean Thm, $[L(x)]^2 = QR^2 + RP^2 = \left( \frac{2+x}{x} \right)^2 + (2+x)^2 = (2+x)^2 \left( {1 + \frac{1}{x^2} } \right)$

c)
$\frac{d[L(x)]^2}{dx} = 2(2+x)\left( {1 + \frac{1}{x^2} } \right) + (2+x)^2 \left( \frac{-2}{x^3} \right) = 2(2+x) \left( {1 + \frac{1}{x^2} - \frac{2+x}{x^3} } \right)$
$= 2(2+x) \left( {1 - \frac{2}{x^3} } \right)$

so, when $x = \sqrt[3] {2}$, that is 0.

d) if we take the second derivative of L(x), we have,
$2(2+x)\left( {\frac{6}{x^4} } \right) + 2\left( {1 - \frac{2}{x^3} } \right)$

and since $x = \sqrt[3] {2}$ is a critical point, we can test this to the second derivative and we have it positive, therefore, $x = \sqrt[3] {2}$is the shortest length..

3. -----------============== 13 ===========---------------
first notice that triangles QRP and BAP are similar because all their angles are the same (why ?), therefore their sides are in the same proportion:

BA / QR = AP / RP

<=> 1 / QR = x / (x+2) ----> QR = (x+2) / x

b. now we simply apply the Pythagorean theorem to the large triangle

[L(x)]^2 = [(x+2)/x]^2 + (2+x)^2 = [(x+2)^2](1 + 1/x^2)

c. differentiate the result from the above clause:

2*(x+2)(1+ 1/x^2) + [(x+2)^2](-2*x^(-3)) = 0

<=> (x+2)(1 + 1/x^2 - 1/x^2 - 2/x^3) = 0
<=> (x+2)(1 - 2/x^3) = 0

now we see that the derivative vanishes at x = -2 and 2^(1/3)
but x=-2 is obviously not physically possible therefore the only solution is x = 2^(1/3).

d. first let us verify that the critical point found earlier is indeed a minimum:

d^2/dx^2 (L(x)) = 2[1 - 2/x^3 + (x+2)(6/x^4)]
d^2/dx^2 (L(x)) |x=2^(1/3) > 0

thus the extrema is indeed a minimum therefore the shortest length is
L(2^(1/3)) ~17 m

4. Thanks for both of your replies, and this might sound stupid, but can you explain this part to me:

$
[L(x)]^2 = QR^2 + RP^2 = \left( \frac{2+x}{x} \right)^2 + (2+x)^2 = (2+x)^2 \left( {1 + \frac{1}{x^2} } \right)
$

I don't get how the two things on either side of the equal sign are the same..

5. notice that:

[(2+x)/x]^2 + (2+x)^2 = [(2+x)^2]*(1 / x^2) + (1)*(2+x)^2

now if we take out the common factor (2+x)^2 we get
[(2+x)^2](1/x^2 + 1)

6. Hello, Fnus!

It took me a few tries to get the first one . . .

12) At 1:00 pm, ship A leaves port $P$ in the direction 30°T at 12 kph.
Also at 1:00 pm, ship B is 100 km east of $P$ and is sailing at 8 kph toward $P.$

Suppose $t$ is the number of hours after 1:00 pm.

(a) Show that the distance $D(t)$ between the two ships is given by:
. . . . $D(t)\;=\;\sqrt{304t^2 - 2800t + 10,000}$

(b) Find the minimum value of $[D(t)]^2$ for all $t > 0.$

(c) At what time, to the nearest minute, are the ships closest?
I assume that "30°T" is the usual bearing.
Code:
      |   6t    A
C * - - - - *
|        *: *
|       * :   *
|      *  :     *
6√3t |  12t*   :       *
|    *    :         *
|30°*     :6√3t       *
|  *      :             *
| *       :               *
|*        :                 *
* - - - - * - - - - - - - - - * - - - - - - *
P         D                   B             Q
: - 6t  - : - - 100-14t - - - : - - 8t  - - :

In $t$ hours, ship A sails at 12 kph from $P$ to $A$: . $PA = 12t$

Since $\Delta ACP$ is a 30-60 right triangle, we have;
. . $CA \,=\, PD \,=\,6t,\;\;CP \,=\, AD \,=\, 6\sqrt{3}t$

Ship B starts from point $Q\!:\;PQ = 100$
In $t$ hours, it sails $8t$ km to $B\!:\;QB = 8t$
Hence: . $PB \:=\:100 - 8t$ and $DB \:=\:(100 - 8t) - 6t \:=\:100-14t$

The distance $D(t)$ between the ships is $AB.$

From right triangle $ADB$, we have: . $D(t) \;=\;AB \;=\;\sqrt{DB^2 + AD^2}$

Hence: . $D(t) \;=\;\sqrt{(100-14t)^2 + (6\sqrt{3}t)^2} \;=\;\sqrt{10,000 - 2800t + 196t^2 + 108t^2}$

Therefore: . ${\color{blue}\boxed{D(t) \;=\;\sqrt{304t^2 - 2800t + 10,000}}}$ . (a)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let $Z \;=\;[D(t)]^2 \;=\;304t^2 - 2800t + 10,000$

Differentiate and equate to zero: . $Z' \;=\;608t - 2800 \:=\:0$

. . $t \:=\:\frac{2800}{608} \:=\:\frac{175}{38}$ hours

The distance is a minimum in $4.605$ hours ≈ 4 hours, 36 minutes.

Therefore, the ships are closest at ${\color{blue}\boxed{5:36\text{ pm.}}}$ .(c)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The desired minimum value is:

. $\left[D\left(\frac{175}{38}\right)\right]^2 \;=\;304\left(\frac{175}{38}\right)^2 - 2800\left(\frac{175}{38}\right) + 10,000 \;=\;{\color{blue}\boxed{\frac{135,000}{19}}}$ .(b)

7. Originally Posted by Fnus
Thanks for both of your replies, and this might sound stupid, but can you explain this part to me:

$
[L(x)]^2 = QR^2 + RP^2 = \left( \frac{2+x}{x} \right)^2 + (2+x)^2 = (2+x)^2 \left( {1 + \frac{1}{x^2} } \right)
$

I don't get how the two things on either side of the equal sign are the same..
okay.. notice that L(x) = PQ right? and you notice that the triangle PRQ is a right triangle whose hypotenuse is PQ. we just applied Pythagorean Theorem there, with legs QR and RP (which are already "given".. the other one was actually solved), and like what "he" said, it is just a matter of factoring.. Ü