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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Maz on December 05, 2015, 06:35:55 pm

Title: Differentiation help plz Unit 3
Post by: Maz on December 05, 2015, 06:35:55 pm: hey- I'm asking a lot of questions these days...
can someone help me with this one plz?
A company has the task of producing spheres of clove 288pi cm^3 +or/and -5. Use differentiation to determine the radius
of the spheres giveng your answer in the form 'a' cm +- 'b' cm, with b rounded to 3 decimal places?
the answer is (6+-0.011) cm
thankyou :)
Title: Re: Differentiation help plz Unit 3
Post by: wyzard on December 07, 2015, 12:47:32 pm: This once again is a small change approximation question, just used in a different way.

It is first crucial to understand what they're asking for. The company likes to produce spheres with volume of $288\pi \quad { cm }^{ 3 }$ , however the volume can exceed or go below this value by $5 \quad { cm }^{ 3 }$ . We usually call this the 'error margin' as in reality we can't produce a sphere of exact volume all the time, there will be some variation, usually expressed as $V\pm \Delta V$ . Our goal in this problem is to work out the the radius and its error margin, expressed as $r\pm \Delta r$

I'll give you pointers to work out the rest. Given the volume of $288\pi \quad { cm }^{ 3 }$ , find the radius first, finding $r$ .

To find the error margin of the radius, this is where small change approximation comes in handy where:

$\Delta r\quad \approx \quad \frac { dr }{ dV } \Delta V$

In this case $\Delta r$ and $\Delta V$ represents the error margin in the small change approximation formula. The reason we can use this is that we are looking at the variation of $r$ when $V$ is allowed to change within its error margin.

Hope this help, finding error margin is just another application of small change approximation in differentiation, pretty cool eh? 8)
Title: Re: Differentiation help plz Unit 3
Post by: Maz on December 07, 2015, 05:26:44 pm: hey- thank you...i got the radius bit but I'm still a bit confused on how to do the
error margin?
Title: Re: Differentiation help plz Unit 3
Post by: wyzard on December 08, 2015, 02:26:12 pm: The error margin of volume is $5{ cm }^{ 3 }$ , which is the $\Delta V$ you're going to use in the small change approximation formula. Same goes for radius.

Think of error margin as the maximum change allowed for volume. Suppose the volume is allowed to exceed $5{ cm }^{ 3 }$ , what is the radius it is allowed to exceed?
Title: Re: Differentiation help plz Unit 3
Post by: Maz on December 08, 2015, 03:59:02 pm: okay cool i got it...thankyou :)