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"If
and be the lengths of a rod of iron at the temperatures t°C and 0°C respectively, then . Find the change of length of the rod per degree Centigrade."So I'm teaching myself calculus from the book Calculus Made Easy by Silvanus Thompson, and this is one of the exercises. I'm pretty sure it would be something like
.I'd love a quick walk through if you can spare the time. I know how to find the differential when I'm working with (simple) equations, but word problems mess me up. Any help is appreciated.
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hi caharris
yes, that's what you want.
The question intends that you treat
as a constant (it is the length at one temperature so is not a variable)
so you just need to differentiate
and multiply by
at the end.
Can you do the differentiation from this and complete the question?
Bob
Last edited by Bob (2011-01-03 08:00:34)
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You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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So I'm basically working with
? Would it be ?Last edited by caharris (2011-01-03 09:17:42)
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Hi caharris;
Welcome to the forum. You did not do that differentiation correctly.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hmmm... I was taught that if
, then .So
, , then ?.?Sorry, I'm still new at this
Thank you guys for helping me out though
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Hi;
You want to differentiate this.
Doing the 2 boxed derivatives one at a time.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Ahh gotcha. Thank you so much!
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Hi caharris;
Your welcome, good luck in your studies. Come back if you need more.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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