Question
The rate of heat loss (in watts) in harbor seal pups has been approximated by$$H(m, T, A)=\frac{15.2 m^{0.67}(T-A)}{10.23 \ln m-10.74}$$where $m$ is the body mass of the pup (in $\mathrm{kg} ),$ and $T$ and $A$ are the body core temperature and ambient water temperature, respectively $\left(\text { in }^{\circ} \mathrm{C}\right) .$ Find the heat loss for the following data. Source: Functional Ecology.a. Body mass $=21 \mathrm{kg} ;$ body core temperature $=36^{\circ} \mathrm{C} ;$ ambient water temperature $=4^{\circ} \mathrm{C}$b. Body mass $=29 \mathrm{kg} ;$ body core temperature $=38^{\circ} \mathrm{C} ;$ ambient water temperature $=16^{\circ} \mathrm{C}$
Step 1
For part a, we have $m = 21$ kg, $T = 36$ degrees Celsius, and $A = 4$ degrees Celsius. Substituting these values into the formula gives us: $$H(21, 36, 4) = \frac{15.2 \times 21^{0.67} \times (36 - 4)}{10.23 \ln 21 - 10.74}$$ Show more…
Show all steps
Your feedback will help us improve your experience
Thomas Emment and 84 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Heat Loss The rate of heat loss (in watts) in harbor seal pups has been approximated by $$H(m, T, A)=\frac{15.2 m^{0.67}(T-A)}{10.23 \ln m-10.74},$$ where $m$ is the body mass of the pup (in kg $),$ and $T$ and $A$ are the body core temperature and ambient water temperature, respectively (in $^{\circ} \mathrm{C} ) .$ Find the heat loss for the following data. Source: Functional Ecology. (a) Body mass $=21 \mathrm{kg} ;$ body core temperature $=36^{\circ} \mathrm{C}$ ambient water temperature $=4^{\circ} \mathrm{C}$ (b) Body mass $=29 \mathrm{kg} ;$ body core temperature $=38^{\circ} \mathrm{C}$ ambient water temperature $=16^{\circ} \mathrm{C}$
Multivariable Calculus
Functions of Several Variables
The rate of heat loss (in watts) in harbor seal pups has been approximated by $$H(m, T, A)=\frac{15.2 m^{0.67}(T-A)}{10.23 \ln m-10.74}$$ where $m$ is the body mass of the pup (in $\mathrm{kg} ),$ and $T$ and $A$ are the body core temperature and ambient water temperature, respectively $\left(\text { in }^{\circ} \mathrm{C}\right) .$ Find the heat loss for the following data. Source: Functional Ecology. a. Body mass $=21 \mathrm{kg} ;$ body core temperature $=36^{\circ} \mathrm{C} ;$ ambient water temperature $=4^{\circ} \mathrm{C}$ b. Body mass $=29 \mathrm{kg} ;$ body core temperature $=38^{\circ} \mathrm{C} ;$ ambient water temperature $=16^{\circ} \mathrm{C}$
Heat Loss In Exercise 56 of Section 2 of this chapter, we found that the rate of heat loss (in watts) in harbor seal pups could be approximated by $$H(m, T, A)=\frac{15.2 m^{0.67}(T-A)}{10.23 \ln m-10.74}$$ where $m$ is the body mass of the pup (in $\mathrm{kg} ),$ and $T$ and $A$ are the body core temperature and ambient water temperature, respectively $\left(\mathrm{in}^{\circ} \mathrm{C}\right) .$ Suppose $m$ is $25 \mathrm{kg}, T$ is $36.0^{\circ},$ and $A$ is $12.0^{\circ} \mathrm{C}$ . Approximate the change in $H$ if $m$ changes to $26 \mathrm{kg}, T$ to $36.5^{\circ},$and $A$ to $10.0^{\circ} \mathrm{C} .$
Total Differentials and Approximations
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD