The safe working load L in pounds for a natural rope can be estimated by L(r) = 5920r^2, where r is the radius of the rope in inches. For an old rope, the function L_0(r) = 4150r^2 is used to estimate its safe working load. What kind of transformation describes this change, and what does this transformation mean?

Question

The safe working load L in pounds for a natural rope can be estimated by L(r) = 5920r^2, where r is the radius of the rope in inches.  For an old rope, the function L_0(r) = 4150r^2 is used to estimate its safe working load. What kind of transformation describes this change, and what does this transformation mean?

anonymous · Answer

@mathmale

anonymous · Answer

I will put it in this format so it will be easier to understand!

The safe working load L in pounds for a natural rope can be estimated by $$L(r) = 5920r^2$$,  where r is the radius of the rope in inches.  For an old rope, the function $$L_0r=4150r^2$$ is used to estimate its safe working load. What kind of transformation describes this change, and what does this transformation mean?

mathmale · Answer

Ricardo:  Compare:
L=5920r^2
L=4150r^2

How did the author obtain the 2nd equation from the first?

perl · Answer

it is a scalar transformation?

mathmale · Answer

R:  Very nice (and professional) presentation!  Thank you.
R:  We talked about two different types of transformations.  Which do you think applies here?

anonymous · Answer

I put them both in a graph to compare them, 

L = 4150r^2 looks to be a bit wider than L = 5920r^2

So what would that mean?

anonymous · Answer

Is it a horizontal transformation?

mathmale · Answer

You're right.  The graph IS wider.  But the proper term here is "vertical stretching/shrinking."

What would the author of these two functions do to the first to obtain the second function?  Can you connect this question with what you and I did before...see parallels?

mathmale · Answer

Hint:
y=x^2
y=cx^2
What's the proper term to describe the transformation of the first to the second?

anonymous · Answer

Ahh, that's right. it is vertical stretching.

perl · Answer

its vertical shrinking

anonymous · Answer

And I am starting to forget some of these terms. I will have to go back to our previous conversations and write them all down.

mathmale · Answer

Right!  In this case, the author multiplied the original equation by a fraction less than one.  So yes, that's within "vertical stretching/shrinking".

perl · Answer

or vertical compression

mathmale · Answer

Yes, that's one thing independent learners have to do.  As you saw earlier this morning, I've already begun forgetting material I used to teach.  So, you'd be doing yourself a big favor by writing out a summary sheet of concepts, vocab., formulas, etc., for each new topic, and review each such sheet regularly.

perl · Answer

and graphically it does stretch a bit horizontally, but for the purpose of this question that is irrelevant. the vertical stretch/shrink is enough

mathmale · Answer

Thank you, perl.  I agree.

perl · Answer

:)

mathmale · Answer

Ricardo:  I'm assuming you're OK with this latest discussion and that we're essentially finished.  Great working with you.  As before, y ou're welcome to share tougher problems with me privately, but that should not stop you from posting them here on Open Study if you wish.  Looking forward to next time!  Hasta la vista, vaya con Dios.

mathmale · Answer

R:  Please type the URL of your home school.  I thought I'd written it down, but no soap.

anonymous · Answer

Okay, that's right! Earlier I had said that because the line had stretched upward, it compressed and came closer together! Aha! I remember that now! Si, gracias! http://fl.okonl.vschoolz.net/

mathmale · Answer

Muchas gracias!  Hasta la vista.