Question

*CHALLENGING QUESTION*

the equation that represents this boundary is y=16sec(pi/36(x))-32, where x and y are both measured in cm. the top of the channel is level with the ground and has a width of 24 cm. the max depth of the channel is 16 cm. find the width of the water surface in the channel when the water depth is 10cm. give answer in form 'a arccos b', where a, b are all reals

Answer 1

diagram with my own additions

Answer 2

@Luigi0210 @wio @dan815 @helpme1.2 anyone? i really need help ;_;

Answer 3

I hate it when my calc is set to degrees :(

This isn't too hard, its just algebra:
set y = -10
\[ -10 = \frac{16}{cos(\frac{\pi}{36}x)}-32 \]
\[ 22cos(\frac{\pi}{36}) = 16 \]
\[ \frac{\pi}{36}x = cos^{-1}(16/22) \]
\[ x= \frac{36}{\pi}cos^{-1}(16/22)\]

this gives you the one x coordinate.
since you are centered at (0,0) the other is just the negative.
so you're width is 2 times that.

\[\boxed{2*\frac{36}{\pi}cos^{-1}(16/22)}\]

Answer 4

ooo finally someone willing to answer!! thank you so much, but this is supposed to be a non calc section D: do you have any idea how to do it without the calc? if not its fine :) thanks for helping!

Answer 5

there is no calculus in my answer

Answer 6

or calculator you mean?

Answer 7

give answer in form 'a arccos b'

Answer 8

yeah no calculator

Answer 9

oh, If you meant my comment it was just me double checking the answer and getting confused why it wasn't working.

The answer in the box is the one you want though.

Answer 10

yeap. thank you so much for answering :D

Answer 11

no problem

Answer 12

I may be wrong but isn't "y" supposed to be set to -6 instead of -10? @polaris_s0i

Answer 13

@lasttccasey i think you may be right because -16+10 is -6; we are counting from the bottom up.

Answer 14

crap, you're right

Answer 15

easy fix:

\[2*\frac{36}{\pi}cos^{-1}(\frac{16}{26})\]

Answer 16

That looks better, which is approximately 20.8081 cm which makes sense when looking at the graph.

Answer 17

good eye :) thanks.

Answer 18

Glad I could help a little.