1. ## Quadratic Functions... and... ranges?

Forgive me if this is in the wrong place. I'm at that transitional stage where the algebra stuff is mix in with the pre-calculus stuff.

The question:

Electric resistance in copper wire changes with the temperature of the wire. If C(t) is the electric resistance at temperature t, in degrees Fahrenheit, then the resistance ratio C(t)/C(0) can be measured.

Temperature t in degrees | C(t)/C(0) ratio
0 | 1
10| 1.0393
20| 1.0798
30| 1.1215
40| 1.1644

(a) (I've already done this step... all it wanted was to find the first-order and second-order differences.)

(b) Find a quadratic formula for the ratio C(t)/C(0) as a function of temperature t.

C(t)/C(0) =

(I used the QuadReg function on my calculator... but I get:

y=ax^2+bx+c
a=6E-6
b=.00387
c=1

I don't know what the E means in a=.)

(c) At what temperature is the electric resistance double that at 0 degrees? (Round your answer to two decimal places.)

____ °

(d) Suppose that you have designed a household appliance to be used at room temperature (69 degrees) and you need to have the wire resistance inside the appliance accurate to plus or minus 8% of the predicted resistance at 69 degrees.

(i) What resistance ratio do you predict at 69 degrees? (Use four decimal places.)

1.2956

(So... the formula my calculator found is supposedly right... but I don't know how to give the formula in a form that my homework can understand.)

(ii) What range of resistance ratios represents plus or minus 8% of the resistance ratio for 69 degrees? (Round your answers to four decimal places.)

____ ± ____

(iii) What temperature range for the appliance will ensure that your appliance operates within the 8% tolerance? (Round your answers to two decimal places.)

____ ° F (smaller value)
____ ° F (larger value)

2. Originally Posted by MathBane
Forgive me if this is in the wrong place. I'm at that transitional stage where the algebra stuff is mix in with the pre-calculus stuff.

The question:

Electric resistance in copper wire changes with the temperature of the wire. If C(t) is the electric resistance at temperature t, in degrees Fahrenheit, then the resistance ratio C(t)/C(0) can be measured.

Temperature t in degrees | C(t)/C(0) ratio
0 | 1
10| 1.0393
20| 1.0798
30| 1.1215
40| 1.1644

(a) (I've already done this step... all it wanted was to find the first-order and second-order differences.)

(b) Find a quadratic formula for the ratio C(t)/C(0) as a function of temperature t.

C(t)/C(0) =

(I used the QuadReg function on my calculator... but I get:

y=ax^2+bx+c
a=6E-6
b=.00387
c=1

I don't know what the E means in a=.)

calculator's way to display scientific notation ... 6E-6 = $\displaystyle \textcolor{red}{6 \times 10^{-6} = 0.000006}$

(c) At what temperature is the electric resistance double that at 0 degrees? (Round your answer to two decimal places.)

____ °

look for the intersection of y = 2 and y = ax^2+bx+c, you should get 197.76 degrees

(d) Suppose that you have designed a household appliance to be used at room temperature (69 degrees) and you need to have the wire resistance inside the appliance accurate to plus or minus 8% of the predicted resistance at 69 degrees.

(i) What resistance ratio do you predict at 69 degrees? (Use four decimal places.)

1.2956

(So... the formula my calculator found is supposedly right... but I don't know how to give the formula in a form that my homework can understand.)

(ii) What range of resistance ratios represents plus or minus 8% of the resistance ratio for 69 degrees? (Round your answers to four decimal places.)

____ ± ____ ... 0.92(1.2956) to 1.08(1.2956)

(iii) What temperature range for the appliance will ensure that your appliance operates within the 8% tolerance? (Round your answers to two decimal places.)

____ ° F (smaller value)
____ ° F (larger value)

set y = the two values found above and calculate the value of x
...