Problem Overview
We calculate the absolute rate d v d T \frac{dv}{dT} d T d v 1 v d v d T \frac{1}{v}\frac{dv}{dT} v 1 d T d v
v ( T ) = ( T 2 + 1 ) ln ( T ) v(T) = (T^2 + 1)^{\ln(T)} v ( T ) = ( T 2 + 1 ) l n ( T )
evaluated at the temperature baseline
T = e T = e T = e .
Step-by-Step Mathematical Solution
Step 1: Formally Apply Logarithmic Differentiation
The governing process is Logarithmic Differentiation :
ln v = ln [ ( T 2 + 1 ) ln T ] \ln v = \ln\left[ (T^2 + 1)^{\ln T} \right] ln v = ln [ ( T 2 + 1 ) l n T ] (This technique is defined in Logarithmic Differentiation .)
Bring the variable exponent down using the Power Property of logarithms:
ln v = ln ( T ) ⋅ ln ( T 2 + 1 ) \ln v = \ln(T) \cdot \ln(T^2 + 1) ln v = ln ( T ) ⋅ ln ( T 2 + 1 ) Step 2: Differentiate Both Sides with Respect to Temperature T T T
Apply the Product Rule on the right side with u = ln ( T ) u = \ln(T) u = ln ( T ) w = ln ( T 2 + 1 ) w = \ln(T^2 + 1) w = ln ( T 2 + 1 )
d d T [ ln v ] = d d T [ ln T ] ⋅ ln ( T 2 + 1 ) + ln ( T ) ⋅ d d T [ ln ( T 2 + 1 ) ] \frac{d}{dT}[\ln v] = \frac{d}{dT}[\ln T] \cdot \ln(T^2 + 1) + \ln(T) \cdot \frac{d}{dT}[\ln(T^2 + 1)] d T d [ ln v ] = d T d [ ln T ] ⋅ ln ( T 2 + 1 ) + ln ( T ) ⋅ d T d [ ln ( T 2 + 1 )] 1 v d v d T = ( 1 T ) ⋅ ln ( T 2 + 1 ) + ln ( T ) ⋅ ( 2 T T 2 + 1 ) \frac{1}{v}\frac{dv}{dT} = \left(\frac{1}{T}\right) \cdot \ln(T^2 + 1) + \ln(T) \cdot \left(\frac{2T}{T^2 + 1}\right) v 1 d T d v = ( T 1 ) ⋅ ln ( T 2 + 1 ) + ln ( T ) ⋅ ( T 2 + 1 2 T ) 1 v d v d T = ln ( T 2 + 1 ) T + 2 T ln T T 2 + 1 \frac{1}{v}\frac{dv}{dT} = \frac{\ln(T^2 + 1)}{T} + \frac{2T \ln T}{T^2 + 1} v 1 d T d v = T ln ( T 2 + 1 ) + T 2 + 1 2 T ln T Step 3: Identify the Fractional Rate of Expansion
Because the problem specifically asks for the fractional rate 1 v d v d T \frac{1}{v}\frac{dv}{dT} v 1 d T d v v v v
Fractional Rate = 1 v d v d T = ln ( T 2 + 1 ) T + 2 T ln T T 2 + 1 \text{Fractional Rate} = \frac{1}{v}\frac{dv}{dT} = \frac{\ln(T^2 + 1)}{T} + \frac{2T \ln T}{T^2 + 1} Fractional Rate = v 1 d T d v = T ln ( T 2 + 1 ) + T 2 + 1 2 T ln T Step 4: Evaluate at T = e T = e T = e
Recall that ln ( e ) = 1 \ln(e) = 1 ln ( e ) = 1
1 v d v d T ∣ T = e = ln ( e 2 + 1 ) e + 2 e ln ( e ) e 2 + 1 = ln ( e 2 + 1 ) e + 2 e e 2 + 1 \left.\frac{1}{v}\frac{dv}{dT}\right|_{T=e} = \frac{\ln(e^2 + 1)}{e} + \frac{2e \ln(e)}{e^2 + 1} = \frac{\ln(e^2 + 1)}{e} + \frac{2e}{e^2 + 1} v 1 d T d v T = e = e ln ( e 2 + 1 ) + e 2 + 1 2 e ln ( e ) = e ln ( e 2 + 1 ) + e 2 + 1 2 e
Numerically, since
e ≈ 2.71828 e \approx 2.71828 e ≈ 2.71828 and
e 2 + 1 ≈ 8.389 e^2 + 1 \approx 8.389 e 2 + 1 ≈ 8.389 :
ln ( 8.389 ) ≈ 2.1269 ⟹ 2.1269 2.7183 + 5.4366 8.389 ≈ 0.7824 + 0.6481 ≈ 1.4305 units − 1 \ln(8.389) \approx 2.1269 \implies \frac{2.1269}{2.7183} + \frac{5.4366}{8.389} \approx 0.7824 + 0.6481 \approx 1.4305\text{ units}^{-1} ln ( 8.389 ) ≈ 2.1269 ⟹ 2.7183 2.1269 + 8.389 5.4366 ≈ 0.7824 + 0.6481 ≈ 1.4305 units − 1 Engineering Conclusion
At T = e T = e T = e ≈ 1.43 K − 1 \approx 1.43\text{ K}^{-1} ≈ 1.43 K − 1 143 % 143\% 143% v ( T ) v(T) v ( T )