Melody

EDIT:  This answer is correct just rediculously long.  I have redone it much more efficiently on a later post.

The two triangles are similar because they each have a right angle and they have a common angle.

Let the third angle be  $$\theta$$

You know that      $$\frac{dx}{dt}=1m/s$$

a) you are being asked to find  $$\frac{dy}{dt}$$  when x=6m

b) you are being asked to find   $$\frac{d(y-x)}{dt}=\frac{dy}{dt}-\frac{dx}{dt}$$

a)  I think it will be easier for me to get y and x in terms of $$\theta$$

$$\begin{array}{rllrll}
tan\theta&=&\frac{y}{6}\qquad \qquad \qquad \qquad & tan\theta&=&\frac{y-x}{1.8}\\\\
y&=&6tan\theta}\qquad& y-x&=&1.8tan\theta\\\\
&&&x&=&y-1.8tan\theta\\\\
&&&x&=&6tan\theta -1.8tan\theta\\\\
&&&x&=&4.2tan\theta \\\\
\frac{dy}{d\theta}&=&6sec^2\theta}\qquad& \frac{dx}{d\theta}&=&4.2sec^2\theta\\\\
\end{array}$$

 ---------------------------------------------------------------------------------------------------------

              $$\begin{array}{rll}
\frac{dy}{dt}&=&\frac{dy}{d\theta}\times \frac{d\theta}{dx}\times \frac{dx}{dt}\\\\
\frac{dy}{dt}&=&6sec^2\theta}\times \dfrac{1}{4.2sec^2\theta}\times 1\\\\
\frac{dy}{dt}&=&\frac{6}{4.2}\\\\
\frac{dy}{dt}&=&\frac{10}{7}\\\\
\frac{dy}{dt}&=&1\frac{3}{7}\;\;m/s\\\\
\end{array}$$

b)  $$\frac{dy}{dt}-\frac{dx}{dt}= 1\frac{3}{7}-1=\frac{3}{7}\;\;m/s$$

So the end of the shadow is moving away at a constant rate of      $$1\frac{3}{7}m/s$$

and

The length of the shadow is changing at a constant rate of     $$\frac{3}{7}\;\;m/s$$

Since both of these are constants, it makes no difference how foar away from the pole the man is.

The answers will always be the same.     

Thanks Rosala, that was a great post.  

I really liked the waterfall in Goa too.  I have a major attraction for waterfalls.

When I was a child there were cliffs and waterfalls at the back of our property.  That is where I liked to play.

I also love dams and all water features.  I love to know where communities get their water supplies from.

Perhaps this is partly because I am Australian and most of our country is desert.  Water is always a big issue.

I really like the trees and all your fun pics too.  Thank you!

Sun 20/7/14

♬                                        ♬ ♬ MELODY ♬ ♬

1) Rosala's Wrap - Thanks Rosala, we really enjoy these. I love the Goa waterfall! And the fun pics as well!

@@ End of Day Wrap :Sat19/7/14   Sydney, Australia   Time 10:30 pm    ♬                                       

It was a quite Fri/Sat today but as always there  were some good answers.  These were provided by CPhill, Alan, Rosala, NinjaDevo, Zegroes, Kitty3 and GoldenLeaf.  Thanks all.        

My new avitar has created some interest.  She is Lady (Queen) Guinevere of Camelot from the Brittish TV series "Merlin".  Arthur was the Legendary King of Camelot.  I was given the title of Queen Guinevere by another member, Morgan Tud.  Morgan Tud was the chief Physician to the royal court of Camelot.  

THANKS NINJA  

I am very sorry Rosala, Ninja is right!  Your answer is PERFECT!   

Thank you Rosala for the translation.

We are all very happy to help you.  Thank you.

------------------------------------------------

Kami sangat senang untuk membantu Anda. Terima kasih.

I am not concerned.  The Queen of Camelot is above such nonsense.

Any peasant who doth not recognise their monach is slow of mind and not worthy of  any consideration.

Umm yes,

every x value only has one y value.

The graph has to pass the straight line test.  This means that any vertical line will never cut the graph more than once.

This might help you.

$50-30c = $49.70

This is an old post that was our "reference material" sticky note topic.

It is worth looking at!