The wind-chill index is a measure of how cold it feels in windy weather, and it is modeled by the...

Question:

The wind-chill index is a measure of how cold it feels in windy weather, and it is modeled by the function

{eq}W(T, v) = 13.12 + 0.6215T - 11.37v^{0.16} + 0.3965Tv^{0.16}, {/eq} where {eq}T {/eq} is the temperature (in {eq}^{\circ} {/eq}C) and {eq}v {/eq} is the wind speed (in km/h).

When {eq}T {/eq} = -5 {eq}^{\circ} {/eq}C and {eq}v {/eq} = 25 km/h, by how much would you expect {eq}W {/eq} to change if:

(a) the actual temperature increases by 1 {eq}^{\circ} {/eq}C?

(b) the wind speed decreases by 1 km/h?

Differentiation:

A mathematical quantity that represents the analyse of the continuous variation of the function with respect to the variable of the function known as differentiation. It used in engineering application to get the result

Answer and Explanation:

.


Given Data:

  • The wind-chill index equation is: {eq}W\left( {T,v} \right) = 13.12 + 0.6215T - 11.37{v^{0.16}} + 0.3965T{v^{0.16}} {/eq}
  • {eq}T = - {5^\circ }{\rm{C}} {/eq}
  • {eq}v = 25\;{\rm{km/h}} {/eq}


(a)

Differentiate the wind-chill index equation with respect to {eq}T {/eq}

{eq}\begin{align*} \dfrac{{d\left( {W\left( {T,v} \right)} \right)}}{{dT}} &= \dfrac{{d\left( {13.12 + 0.6215T - 11.37{v^{0.16}} + 0.3965T{v^{0.16}}} \right)}}{{dT}}\\ &= 0.6215 + 0.3965{v^{0.16}} \end{align*} {/eq}


Substitute the value and solve the above expression {eq}v = 25\;{\rm{km/h}} {/eq}

{eq}\begin{align*} \dfrac{{d\left( {W\left( {T,v} \right)} \right)}}{{dT}} &= 0.6215 + 0.3965{\left( {25} \right)^{0.16}}\\ &= 0.6215 + 0.6636\\ &= 1.2851 \end{align*} {/eq}


Thus the {eq}W {/eq} changes is {eq}1.2851 {/eq} if actual temperature increased by {eq}{1^\circ }{\rm{C}} {/eq}


(b)

Partial differentiate the wind-chill index equation with respect to {eq}T {/eq}

{eq}\begin{align*} \dfrac{{d\left( {W\left( {T,v} \right)} \right)}}{{dT}} &= \dfrac{{d\left( {13.12 + 0.6215T - 11.37{v^{0.16}} + 0.3965T{v^{0.16}}} \right)}}{{dT}}\\ &= 0.6215 - 11.37\left( {0.16} \right){v^{ - 0.84}}\dfrac{{dv}}{{dT}} + 0.3965{v^{0.16}} + 0.16\left( {0.3965} \right){v^{ - 0.84}}\dfrac{{dv}}{{dT}} \end{align*} {/eq}


Substitute the value and solve the above expression at {eq}v = 25\;{\rm{km/h}} {/eq} and {eq}\dfrac{{dv}}{{dT}} = - 1\;{\rm{km/hr}} {/eq}

{eq}\begin{align*} \dfrac{{d\left( {W\left( {T,v} \right)} \right)}}{{dT}} &= 0.6215 - 11.37\left( {0.16} \right){\left( {25} \right)^{ - 0.84}}\left( { - 1} \right) + 0.3965{\left( {25} \right)^{0.16}} + 0.16\left( {0.3965} \right){\left( {25} \right)^{ - 0.84}}\left( { - 1} \right)\\ &= 0.6215 - 0.1217 + 0.6636 - 0.00424\\ &= 1.1591 \end{align*} {/eq}


Thus the {eq}W {/eq} changes is {eq}1.1591 {/eq} if wind speed decrease by {eq}1\;{\rm{km/h}} {/eq}


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Basic Calculus: Rules & Formulas

from Calculus: Tutoring Solution

Chapter 3 / Lesson 6
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