Answer to Question #253704 in Calculus for barbie

Question

In the temperature range between 0

0C and 7000C the resistance R [in ohms] of a certain platinum resistance

thermometer is given by

R = 10 + 0.04124T − 1.779 × 10−5T

2

where T is the temperature in degrees Celsius. Where in the interval from 0

0C to 7000C is the resistance of the

thermometer most sensitive and least sensitive to temperature changes? [Hint: Consider the size of dR/dT in the

interval 0 ≤ T ≤ 700.].

Answer 1

The derivative of the function measures the rate at which function changes with respect to the change of the variable. So, to find where the given function is more/less sensitive to the changes of time we should find maximum/minimum of its derivation.

$R = 10 + 0.04124T − 1.779 × 10^{−5T^2}$

${\frac {dR} {dT}}=0.04124+17.79*ln(10)*T*10^{-5t^{2}}$

Now to find extremums of this function we must find it's derivative

${\frac {d^{2}R} {dT^{2}}}=17.19*ln(10)*10^{-5t^{2}}-177.9*ln^{2}(10)*t^{2}*10^{-5t^{2}}$

Next step is to find the critical points:

$17.19*ln(10)*10^{-5t^{2}}-177.9*ln^{2}(10)*t^{2}*10^{-5t^{2}}=0$

$17.19*ln(10)*10^{-5t^{2}}(1-10*ln(10)*t^{2})=0$

$10^{-5t^{2}}(1-10*ln(10)*t^{2})=0$

Since $10^{-5t^{2}}$is never equal to 0, then

$1-10*ln(10)*t^{2}=0\to t = {\frac 1 {\sqrt{10*ln(10)}}}$

We have found the critical points, now we should verify the max and min at t є [0;700]

Let ${\frac {dR} {dT}}=f(t)$

$f(0) = 0.04124$

$f({\frac 1 {\sqrt{10*ln(10)}}})= 5.191$

$f(700) = 0.04124 + 17.79+ln(10)*700*10^{-5*700^{2}}$

$7.79+ln(10)*700*10^{-5*700^{2}}$ is a bit higher than 0, than

$f(0) < f(700) < f({\frac 1 {\sqrt{10*ln(10)}}})$

The most sensitive thermometer is at point $T = {\frac 1 {\sqrt{10*ln(10)}}}$ , less sensitive is at point T = 0