Questions: The following table shows the wind chill C(T, w) (in °F) as a function of the air temperature T (in °F) and the wind speed w (in miles per hour) according to the current National Weather Service model. 5 10 15 20 25 30 35 ------------------------ 40 36 34 32 30 29 28 28 35 31 27 25 24 23 22 21 30 25 21 19 17 16 15 14 T 19 15 13 11 9 8 7 25 13 9 6 4 3 1 0 15 7 3 0 -2 -4 -5 -7 10 1 -4 -7 -9 -11-12-14 There are three ways to estimate a partial derivative from a table of values, the forward difference quotient (for example, fx(a, b) approximates (f(a+h, b)-f(a, b))/h with h>0), the backward difference quotient (for example, fx(a, b) approximates (f(a, b)-f(a-h, b))/h with h>0), and the symmetric difference quotient (for example, fz(a, b) approximates (f(a+h, b)-f(a-h, b))/(2 h) with h>0). The symmetric difference quotient is just the average of the forward and backward difference quotients. The forward difference quotient approximation to CT(15,25) is °F / °F. The backward difference quotient approximation to CT(15,25) is °F / °F. The symmetric difference quotient approximation to CT(15,25) is °F / °F. The forward difference quotient approximation to Cw(15,25) is -0.2 °F / mph. The backward difference quotient approximation to Cw(15,25) is -0.4 °F / mph. The symmetric difference quotient approximation to Cw(15,25) is °F / mph. The National Weather Service has also provided an empirical formula for approximating the wind chill. It states that C(T, w) approximates 35.74+0.6215 T+w^(0.10)(-35.75+0.4275 T) Based on this formula, CT(15,25) approximates °F / °F and Cw(15,25) approximates °F / mph.

The following table shows the wind chill C(T, w) (in °F) as a function of the air temperature T (in °F) and the wind speed w (in miles per hour) according to the current National Weather Service model. 5 10 15 20 25 30 35 ------------------------ 40 36 34 32 30 29 28 28 35 31 27 25 24 23 22 21 30 25 21 19 17 16 15 14 T 19 15 13 11 9 8 7 25 13 9 6 4 3 1 0 15 7 3 0 -2 -4 -5 -7 10 1 -4 -7 -9 -11-12-14 There are three ways to estimate a partial derivative from a table of values, the forward difference quotient (for example, fx(a, b) approximates (f(a+h, b)-f(a, b))/h with h>0), the backward difference quotient (for example, fx(a, b) approximates (f(a, b)-f(a-h, b))/h with h>0), and the symmetric difference quotient (for example, fz(a, b) approximates (f(a+h, b)-f(a-h, b))/(2 h) with h>0). The symmetric difference quotient is just the average of the forward and backward difference quotients. The forward difference quotient approximation to CT(15,25) is °F / °F. The backward difference quotient approximation to CT(15,25) is °F / °F. The symmetric difference quotient approximation to CT(15,25) is °F / °F. The forward difference quotient approximation to Cw(15,25) is -0.2 °F / mph. The backward difference quotient approximation to Cw(15,25) is -0.4 °F / mph. The symmetric difference quotient approximation to Cw(15,25) is °F / mph. The National Weather Service has also provided an empirical formula for approximating the wind chill. It states that C(T, w) approximates 35.74+0.6215 T+w^(0.10)(-35.75+0.4275 T) Based on this formula, CT(15,25) approximates °F / °F and Cw(15,25) approximates °F / mph.
Transcript text: The following table shows the wind chill $C(T, w)$ (in $\left.{ }^{\circ} F\right)$ as a function of the air temperature $T$ (in ${ }^{\circ} F$ ) and the wind speed $w$ (in miles per hour) according to the current National Weather Service model. \begin{tabular}{rrrrrrrr} \\ & 5 & 10 & 15 & 20 & 25 & 30 & 35 \\ 40 & 36 & 34 & 32 & 30 & 29 & 28 & 28 \\ 35 & 31 & 27 & 25 & 24 & 23 & 22 & 21 \\ 30 & 25 & 21 & 19 & 17 & 16 & 15 & 14 \\ T & 19 & 15 & 13 & 11 & 9 & 8 & 7 \\ 25 & 13 & 9 & 6 & 4 & 3 & 1 & 0 \\ 15 & 7 & 3 & 0 & -2 & -4 & -5 & -7 \\ 10 & 1 & -4 & -7 & -9 & -11 & -12 & -14 \end{tabular} There are three ways to estimate a partial derivative from a table of values, the forward difference quotient (for example, $f_{x}(a, b) \approx \frac{f(a+h, b)-f(a, b)}{h}$ with $\left.h>0\right)$, the backward difference quotient (for example, $f_{x}(a, b) \approx \frac{f(a, b)-f(a-h, b)}{h}$ with $h>0$ ), and the symmetric difference quotient (for example, $f_{z}(a, b) \approx \frac{f(a+h, b)-f(a-h, b)}{2 h}$ with $\left.h>0\right)$. The symmetric difference quotient is just the average of the forward and backward difference quotients. The forward difference quotient approximation to $C_{T}(15,25)$ is $\square$ ${ }^{\circ} F /{ }^{\circ} F$. The backward difference quotient approximation to $C_{T}(15,25)$ is $\square$ ${ }^{\circ} \mathrm{F} /{ }^{\circ} \mathrm{F}$. The symmetric difference quotient approximation to $C_{T}(15,25)$ is $\square$ ${ }^{\circ} F /{ }^{\circ} F$. The forward difference quotient approximation to $C_{w}(15,25)$ is $-0.2 \quad \circ / \mathrm{mph}$. The backward difference quotient approximation to $C_{w}(15,25)$ is $\square$ $-0.4$ ${ }^{\circ} F / \mathrm{mph}$. The symmetric difference quotient approximation to $C_{w}(15,25)$ is $\square$ ${ }^{\circ} F / \mathrm{mph}$. The National Weather Service has also provided an empirical formula for approximating the wind chill. It states that \[ C(T, w) \approx 35.74+0.6215 T+w^{010}(-35.75+0.4275 T) \] Based on this formula, $C_{T}(15,25) \approx \square$ $\square$ ${ }^{\circ} F /{ }^{\circ} F$ and $C_{w}(15,25) \approx$ $\square$ ${ }^{\circ} \mathrm{F} / \mathrm{mph}$.
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Solution

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Solution Steps

To solve the problem, we need to calculate the partial derivatives of the wind chill function C(T,w) C(T, w) using the given table and the empirical formula. For the table, we will use the forward, backward, and symmetric difference quotients to approximate the partial derivatives CT(15,25) C_T(15, 25) and Cw(15,25) C_w(15, 25) . For the empirical formula, we will compute the partial derivatives analytically.

  1. Forward Difference Quotient for CT(15,25) C_T(15, 25) : Use the values from the table for T=15 T = 15 and T=25 T = 25 with w=25 w = 25 .
  2. Backward Difference Quotient for CT(15,25) C_T(15, 25) : Use the values from the table for T=15 T = 15 and T=10 T = 10 with w=25 w = 25 .
  3. Symmetric Difference Quotient for CT(15,25) C_T(15, 25) : Average the forward and backward difference quotients.
  4. Forward Difference Quotient for Cw(15,25) C_w(15, 25) : Use the values from the table for w=25 w = 25 and w=30 w = 30 with T=15 T = 15 .
  5. Backward Difference Quotient for Cw(15,25) C_w(15, 25) : Use the values from the table for w=25 w = 25 and w=20 w = 20 with T=15 T = 15 .
  6. Symmetric Difference Quotient for Cw(15,25) C_w(15, 25) : Average the forward and backward difference quotients.
  7. Empirical Formula Partial Derivatives: Differentiate the empirical formula with respect to T T and w w and evaluate at T=15 T = 15 and w=25 w = 25 .
Step 1: Calculate Forward Difference Quotient for CT(15,25) C_T(15, 25)

Using the values from the table, the forward difference quotient is calculated as follows: CT(15,25)≈C(25,25)−C(15,25)25−15=3−(−4)10=710=0.7 C_T(15, 25) \approx \frac{C(25, 25) - C(15, 25)}{25 - 15} = \frac{3 - (-4)}{10} = \frac{7}{10} = 0.7

Step 2: Calculate Backward Difference Quotient for CT(15,25) C_T(15, 25)

The backward difference quotient is calculated using: CT(15,25)≈C(15,25)−C(10,25)15−10=−4−(−5)5=15=0.2 C_T(15, 25) \approx \frac{C(15, 25) - C(10, 25)}{15 - 10} = \frac{-4 - (-5)}{5} = \frac{1}{5} = 0.2

Step 3: Calculate Symmetric Difference Quotient for CT(15,25) C_T(15, 25)

The symmetric difference quotient is the average of the forward and backward quotients: CT(15,25)≈0.7+0.22=0.92=0.45 C_T(15, 25) \approx \frac{0.7 + 0.2}{2} = \frac{0.9}{2} = 0.45

Step 4: Calculate Forward Difference Quotient for Cw(15,25) C_w(15, 25)

For the wind speed, the forward difference quotient is: Cw(15,25)≈C(15,30)−C(15,25)30−25=−5−(−4)5=−15=−0.2 C_w(15, 25) \approx \frac{C(15, 30) - C(15, 25)}{30 - 25} = \frac{-5 - (-4)}{5} = \frac{-1}{5} = -0.2

Step 5: Calculate Backward Difference Quotient for Cw(15,25) C_w(15, 25)

The backward difference quotient is calculated as: Cw(15,25)≈C(15,25)−C(15,20)25−20=−4−(−2)5=−25=−0.4 C_w(15, 25) \approx \frac{C(15, 25) - C(15, 20)}{25 - 20} = \frac{-4 - (-2)}{5} = \frac{-2}{5} = -0.4

Step 6: Calculate Symmetric Difference Quotient for Cw(15,25) C_w(15, 25)

The symmetric difference quotient is: Cw(15,25)≈−0.2+(−0.4)2=−0.62=−0.3 C_w(15, 25) \approx \frac{-0.2 + (-0.4)}{2} = \frac{-0.6}{2} = -0.3

Step 7: Evaluate Empirical Formula for CT(15,25) C_T(15, 25) and Cw(15,25) C_w(15, 25)

Using the empirical formula: CT(15,25)≈1.2113 C_T(15, 25) \approx 1.2113 Cw(15,25)≈−0.1619 C_w(15, 25) \approx -0.1619

Final Answer

The results for the partial derivatives are:

  • CT(15,25) C_T(15, 25) (Forward): 0.7 0.7

  • CT(15,25) C_T(15, 25) (Backward): 0.2 0.2

  • CT(15,25) C_T(15, 25) (Symmetric): 0.45 0.45

  • Cw(15,25) C_w(15, 25) (Forward): −0.2 -0.2

  • Cw(15,25) C_w(15, 25) (Backward): −0.4 -0.4

  • Cw(15,25) C_w(15, 25) (Symmetric): −0.3 -0.3

  • Empirical CT(15,25) C_T(15, 25) : 1.2113 1.2113

  • Empirical Cw(15,25) C_w(15, 25) : −0.1619 -0.1619

Thus, the final boxed answers are: CT(15,25) (Forward)=0.7,CT(15,25) (Backward)=0.2,CT(15,25) (Symmetric)=0.45 \boxed{C_T(15, 25) \text{ (Forward)} = 0.7, \quad C_T(15, 25) \text{ (Backward)} = 0.2, \quad C_T(15, 25) \text{ (Symmetric)} = 0.45} Cw(15,25) (Forward)=−0.2,Cw(15,25) (Backward)=−0.4,Cw(15,25) (Symmetric)=−0.3 \boxed{C_w(15, 25) \text{ (Forward)} = -0.2, \quad C_w(15, 25) \text{ (Backward)} = -0.4, \quad C_w(15, 25) \text{ (Symmetric)} = -0.3} CT(15,25) (Empirical)=1.2113,Cw(15,25) (Empirical)=−0.1619 \boxed{C_T(15, 25) \text{ (Empirical)} = 1.2113, \quad C_w(15, 25) \text{ (Empirical)} = -0.1619}

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