Determining the rate of change of the function

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andym1512

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Hi guys, ive got an interesting engineering related calculus problem and im struggling with it. Any help would be much appreciated guys, thank you!

The legnth of a metal rod L metres at temperature £ oC is given by L = 1 +0.00003£ + 0.0000006£2. Determine the rate of change of L in mm/oC when the temperature is 75 oC.

Apologies for using the '£' sign, its just a symbol for a variable. Not sure how to use the correct math symbol when posting on her :s
 
andym1512 said:
Hi guys, ive got an interesting engineering related calculus problem and im struggling with it. Any help would be much appreciated guys, thank you!

The legnth of a metal rod L metres at temperature £ oC is given by L = 1 +0.00003£ + 0.0000006£2. Determine the rate of change of L in mm/oC when the temperature is 75 oC.

Apologies for using the '£' sign, its just a symbol for a variable. Not sure how to use the correct math symbol when posting on her :s
Click to expand...
If you are struggling, I hope it is because you are NOT taking calculus at the moment, but rather have been given this question in a physics or engineering course.

The derivatives of all three terms are simple.
1st term, derivative of a constant is 0
2nd term, derivative of 1st power is 1
3rd term, derivative of x^2 is 2x
Conversion from m to mm, multiply by 1000

dL/dt = 0.03 mm/C + (0.0012 mm/C^2)*(75 C) = 0.12 mm/C
 
How do you get to the answer 0.12 from those workings?

I have the same question and work it out as follows -

L = 1 + 0.00003T + 0.0000006T^2

DL/DT = 0 + 0T + 2 x 0.0000006T (x 1000 for mm)

DL/DT = 0.012


Please correct me?
 
dave turbo said:
How do you get to the answer 0.12 from those workings?

I have the same question and work it out as follows -

L = 1 + 0.00003T + 0.0000006T^2

DL/DT = 0 + 0T + 2 x 0.0000006T (x 1000 for mm)
Click to expand...
No, dL/dT= 0+ 0.00003+ 2 x 0.0000006T= 0.00003+ .0000012T
Set T= 75: 0.00003+ 0.00009= 0.00012, NOT 0.012

DL/DT = 0.012


Please correct me?
Click to expand...
 
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