# A bit of fun homework as always.. tides

• Nov 20th 2006, 08:31 AM
rom
A bit of fun homework as always.. tides
:) :confused: hi all, just need a bit of guidence with the following.. i don't know where to begin..
so if any1 can help i would really appreciate it!!

1. Entering the harbour

The depth y in metres above a horizontal mark on a harbour wall is given by
y=3sin(pi/12)
?
where t? is the time in hours after midday on Monday.

(a) Plot the graph of y=3sin(pi/12) ? for ? 0_< t _< 36.
Hope you cn understand the notation.. this one mean 0 is equal or less than t etc..

(b) The rate of change of depth of water in the harbour is given by dy/dt ? Find dy/dt ? and calculate the rate of change of depth of water when t=19?

(c) List the coordinates of all the stationary points on the graph.

(d) A luxury liner can only enter the harbour when the depth of water is at least 1.5 metres above the horizontal mark. At what times during the week can the liner enter the harbour?

ps. ignore those question marks...not sure what they are there for????
• Nov 20th 2006, 10:26 AM
topsquark
Quote:

Originally Posted by rom
y=3sin(pi/12)

As there is no "t" in the given equation, these questions cannot be answered.

-Dan
• Nov 20th 2006, 01:07 PM
Soroban
Hello, rom!

I will assume that a $t$ was omitted from the function.

Quote:

1. Entering the harbour

The depth $y$ in metres above a horizontal mark on a harbour wall is given by:
. . $y \:=\:3\sin\left(\frac{\pi}{12}t\right)$ .where $t$ is the time in hours after midday on Monday.

(a) Plot the graph of $y \:=\:3\sin\left(\frac{\pi}{12}t\right)$ for $0 \leq t \leq 36$

(b) The rate of change of depth of water in the harbour is given by $\frac{dy}{dt}$
Find $\frac{dy}{dt}$and calculate the rate of change of depth of water when $t = 19$

(c) List the coordinates of all the stationary points on the graph.

(d) A luxury liner can only enter the harbour when the depth of water
is at least 1.5 metres above the horizontal mark.
At what times during the week can the liner enter the harbour?

(a) The graph looks like this:
Code:

        |       3 +      *                              *         |  *  :  *                      *  :  *         | *    :    *                  *    :    *         |*      :      *                *      :      *         |      :              18              :     ---*-------+-------*-------+-------*-------+-------*---         |      6      12      :      24      30      36         |                *      :      *         |                *    :    *         |                  *  :  *       -3+                      *         |

(b) The derivative is: . $\frac{dy}{dt}\:=\:3\cos\left(\frac{\pi}{12}t\right )\cdot\frac{\pi}{12} \:=\:\frac{\pi}{4}\cos\left(\frac{\pi}{12}t\right)$

When $t = 19,\;\frac{dy}{dt} \:=\:\frac{\pi}{4}\cos\left(\frac{19\pi}{12}\right ) \:\approx\: 0.2$ metres per hour (rising).

(c) On the graph, we see that the startionary points (horizontal tangents):
. . are at: . $(6,\,3),\:(18,\,-3),\:(30,\,3)$

(d) When is $y \geq 1.5\,?$

The first time $y = 1.5$, we have: . $3\sin\left(\frac{\pi}{12}t\right) \:=\:1.5\quad\Rightarrow\quad \sin\left(\frac{\pi}{12}t\right) \:=\:0.5$
. . Then: . $\frac{\pi}{12}t\:=\:\frac{\pi}{6}\quad\Rightarrow\ quad t = 2$

The next time $y = 1.5$ is $t = 10$

On the graph, we see that $y \geq 1.5$ for: $2 \leq t \leq 10$
. . and it happens again for: $26 \leq t \leq 34$

Therefore, the liner can enter the harbour on Monday from 2 pm to 10 pm
. . and on Tuesday from 2 pm to 10 pm.

• Nov 20th 2006, 01:39 PM
rom
Thank you!! but....
hi soroban!! i have just looked back at the equation, and the equation is

y=3sin( pi t/12)

so its y= 3 sin (pi x t over 12)

So does this change the results much or in any way?? Really sorry to let you know this now!! you did so much work!!! i do really appreciate your help tho!!

if you can help further.....