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Differentiation Problem! - HSN forum

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Differentiation Problem!


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#1 louisa567

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Posted 08 October 2006 - 11:45 AM

Our teacher has asked us to do Q4 and 5 on differentiation pg 115, Ex 6R.

4. A ship has a 200 Km journey to make at a constant speed. At x km/h the cost of the journey will be

(x squared + 4000/x) per hour.

a. Find an expression for the time taken for the journey
the total cost of the trip.

Im completely lost with part a.

b. Find the speed that minimises the cost of the journey and calculate this minimum cost.

I got as far to find x = 12.6. Is this right?

However when I do the table I get a max value rather than a minimum value.

Ive got the same problem with Q5 a and b. Any ideas?

#2 John

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Posted 08 October 2006 - 12:45 PM

Is the equation

(x power2.gif + 4000)/x

or

x power2.gif + (4000/x)

#3 louisa567

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Posted 08 October 2006 - 05:07 PM

Its (x2 + 4000/x) - Its all inside one bracket. Just the 4000 is divided by x.


#4 Steve

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Posted 08 October 2006 - 05:30 PM

For part (a), you have to remember that \textrm{speed}=\displaystyle\frac{\textrm{distance}}{\textrm{time}}. It's just a case of rearranging and substituting.

For (b), you want to differentiate the cost and set equal to zero. Solving for x gives you the stationary values. When differentiating x^2+\displaystyle\frac{4000}{x}, you need to write the second term with a negative power first.

You might have differentiated wrongly. I got x=\sqrt[3]{2000}.

Hope this helps.
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#5 louisa567

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Posted 08 October 2006 - 05:39 PM

Thanks Steve. So the next line to differentiate would be x cubed + 4000

Maybe not!

#6 Steve

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Posted 08 October 2006 - 05:49 PM

No, that's not right. You should have



\begin{align*}x^2+\displaystyle\frac{4000}{x}=x^2+4000x^{-1}\end{align*}

I see what you've done, you've multiplied everything by x, is that right?

The reason you can't do that is because



\begin{align*}x^2+\displaystyle\frac{4000}{x}=x^3+4000\end{align*}

is not an correct equation, the two things are not equal (think about their graphs). You can only put an equals symbol between two things that have the same value!

This is like saying that if the temperature of water after being heated is given by \frac1x then then temperature is given by "1", which it's clearly not!

You see?
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#7 louisa567

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Posted 08 October 2006 - 05:54 PM

Yea thanks for that Steve. That makes sense now.

Ive got as far as x = 3rd root 2000 like you had. But when I sub this into (2x - (4000/ x squared)) it doesnt seem to work.

#8 Steve

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Posted 08 October 2006 - 05:55 PM

Are you still having trouble with Q5?
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#9 louisa567

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Posted 08 October 2006 - 05:57 PM

Im getting a max value rather than a min one.

Yea 5 is just as hard.

#10 Steve

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Posted 08 October 2006 - 06:57 PM

For Q4, x=\sqrt[3]{2000}\approx12.6 is definitely a minimum. Suppose the cost is C.

When x<12.6, say x=12, \frac{dC}{dx}=2\times12-\frac{4000}{12^2}\approx -3.8<0.
When x>12.6, say x=13, \frac{dC}{dx}=2\times13-\frac{4000}{13^2}\approx2.3>0.

This gives:



\begin{align*}
\begin{array}{*{20}c}
   x &\vline &   \to  & {\sqrt[3]{{2000}}} &  \to   \\
\hline
   {\frac{{dC}}{{dx}}} &\vline &   -  & 0 &  +   \\
   {{\rm Graph}} &\vline &  \backslash  & \_ & /  \\
\end{array}
\end{align*}

For Q5, you know that the volume V=144 cm3. But you also know a way to work this volume out.

The volume of the cylinder is given by \textrm{area of base}\times\textrm{height}, and whatever this is must be equal to 144.

Does that help?
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#11 louisa567

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Posted 08 October 2006 - 07:00 PM

Oh so then to find the speed and minimum cost is it just a case of speed, distance, time. The answer at the back of the book is 20 km/h and 6000. This could be wrong though.

Yea that helps Steve - you are a maths genius!

What do they sub into what?

#12 louisa567

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Posted 08 October 2006 - 07:20 PM

Hopefully it wont get much harder than this!

#13 Steve

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Posted 08 October 2006 - 07:22 PM

I'm so sorry, I didn't read the question properly. Most of what I've told you is wrong (well not wrong but not relevant for this question).

The time taken is just given by \frac{\textrm{distance}}{\textrm{speed}}, both of which you are given (the distance is 200 and the speed is x).

Then the cost is just the time multiplied by the cost per hour. I get that this is 200x-800000x^{-2}. Differentiating and solving equal to zero gives you the answer in the book. Then all you have to do is put this speed into your formula for cost.

I'm really sorry, I stupidly took the given equation as the total cost. Oh well, at least you've had lots of practice at differentiating!
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#14 louisa567

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Posted 08 October 2006 - 07:32 PM

Lol thanks for your help. I should have noticed really. Thanks for all your help !!!

So 4a. T = D/S = 200/x

How do you get the cost per hour for the second part?


#15 Steve

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Posted 08 October 2006 - 07:33 PM

You're given that in the question.
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#16 louisa567

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Posted 08 October 2006 - 07:42 PM

So thats just (x squared + (4000/x)) x 200/x

Im still not sure!

So what is the working for part b? I cant do part 2 of a so how would you begin with the second part?

#17 Steve

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Posted 08 October 2006 - 08:27 PM

Yes, you're right. That gives you the total cost for the journey (that's (a)(ii) done!). It's this that you have to work with in part (b). So you have to write x^2+\frac{4000}{x} as one fraction then do the multiplication, then differentiate etc.

You should end up with what I said above.
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#18 louisa567

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Posted 09 October 2006 - 09:04 PM

Thanks for your help Steve. No one our class really got it but now I do!!!





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