Integral/Derivative Problem Solving Question

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Code: 
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| Distance x in cm |   0 |  1 |  5 |  6 |  8 |
-------------------|-----|----|----|----|----|
| Temp t(x) in °C  | 100 | 93 | 70 | 62 | 55 |
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Metal wire is 8 cm. and heated at one end. Distance x is how far from the heated end you are. The function t is decreasing and twice differentiable.

a) Estimate t'(7)
b) Write an integral expression of T(x) for average temp of wire. Estimate average temp of wire using trapezoidal sum with 4 subintervals indicated by data.
c) Find integral from 0 to 8. Explain the meaning of this in the context of the problem.
d) Is that data consistent with the assertion that T''(x) > 0 for every x from 0 to 8? Explain.
 
(a) slope of a secant line is the best you can do to approximate the slope of a tangent line ... use the values in the table wisely
t'(x) is approx [t(x<sub>2</sub>) - t(x<sub>1</sub>)]/(x<sub>2</sub> - x<sub>1</sub>)

(b) Favg = 1/(b-a)*INT{a to b} F(x) dx

trap sum = (delta x)/(2n)*[f(x<sub>0</sub>) + 2f(x<sub>1</sub>) + 2f(x<sub>2</sub>) + ... + 2f(x<sub>n-1</sub>) + f(x<sub>n</sub>)]

(c) Look at the statement of this question again ... you can only estimate the integral of T(x) with a table of data. To understand the meaning, look at the units you get from integrating T(x).

(d) what does T"(x) > 0 say about the curve T(x)?
 
(a) slope of a secant line is the best you can do to approximate the slope of a tangent line ... use the values in the table wisely
t'(x) is approx [t(x<sub>2</sub>) - t(x<sub>1</sub>)]/(x<sub>2</sub> - x<sub>1</sub>)

(b) Favg = 1/(b-a)*INT{a to b} F(x) dx

trap sum = (delta x)/(2n)*[f(x<sub>0</sub>) + 2f(x<sub>1</sub>) + 2f(x<sub>2</sub>) + ... + 2f(x<sub>n-1</sub>) + f(x<sub>n</sub>)]

(c) Look at the statement of this question again ... you can only estimate the integral of T(x) with a table of data. To understand the meaning, look at the units you get from integrating T(x).

(d) what does T"(x) > 0 say about the curve T(x)?
Click to expand...

For C and D, i dont get the work you are giving there - the table is on the top of my questions btw
 
calcuman said:
For C and D, i dont get the work you are giving there - the table is on the top of my questions btw
Click to expand...

I see the table.

part C asks to find the integral of T(x) from 0 to 8 ... I'm telling you that an integral can only be estimated from the table of data. So, recheck the question and confirm what is actually being asked for.

part D ... look at how the average slope changes between each consecutive data point ... what is happening to the slopes as you go from 0 to 1, 1 to 5, 5 to 6, and 6 to 8? Now, you tell me, is that consistent with the claim that T"(x) > 0?
 
Part c actually was asking for the Integral, a=0, b=0 of T'(x) not T(x)
 
Also for Part A, did you find the slope AKA T'(7) by f(x2)-f(x1)/x2-x1? I'm just confused by the notation
 
calcuman said:
Also for Part A, did you find the slope AKA T'(7) by f(x2)-f(x1)/x2-x1? I'm just confused by the notation
Click to expand...

that is the general formula to find the slope between two points, isn't it?
you are supposed to estimate T'(7) (the slope of a tangent line) by using the slope of a secant line ... now, what two values in the table would you use to do that?

calcuman said:
Part c actually was asking for the Integral, a=0, b=0 of T'(x) not T(x)
Click to expand...

Is that so? Big difference from what you previously asked.

hint for part C: INT{a to b} f'(x) dx = ?
ever heard of the Fundamental Theorem of Calculus?
 
yes it equals T(x) in which I just subtract T(8)-T(0)= -45
but what i'm not sure if what the units of measurement is supposed to be? I would think CM/C in that its the change in temperature with repsect to the distance
 
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