Derivative Word Problem Help

kss

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At 0 degrees Celsius, the heat loss H (in Kilocalories per square meter for hour) from a person's body can be modeled by

H = 33(10(sqrt(v)) - v + 10.5)

Where v is the wind speed (in meters per second).


A)Find dH/dv, and interpret its meaning in this situation.

dH/dv = 33(5/(sqrt(v)) - 1)

I have the derivative but I'm having a hard time trying to interpret its meaning???


B) Find the rate of change of H when v=5.

Also do I just plug in 5 into the derivative I got to find the rate of change?



If anyone could help me I would really appreciate it... been going through textbook for hours trying to figure this out...
 
At 0 degrees Celsius, the heat loss H (in Kilocalories per square meter for hour) from a person's body can be modeled by

H = 33(10(sqrt(v)) - v + 10.5)

Where v is the wind speed (in meters per second).


A)Find dH/dv, and interpret its meaning in this situation.

dH/dv = 33(5/(sqrt(v)) - 1)

I have the derivative but I'm having a hard time trying to interpret its meaning???
The only thing I could think to say is that the rate of change of H, with respect to v, is inversely proportional to the square root of v.

Find the rate of change of H when v=5.
Also do I just plug in 5 into the derivative I got to find the rate of change?

Yes, that's all.


If anyone could help me I would really appreciate it... been going through textbook for hours trying to figure this out...
 
Find dH/dv, and interpret its meaning in this situation.

This is like dy/dx -- the rate at which variable y changes with respect to variable x. Only, in this scenario, y is called H and x is called v. H is a function of v, and dH/dv is the derivative of H with respect to v.

Symbol dH/dv represents the rate at which H changes, with respect to v. Therefore, dH/dv can be interpreted as the rate at which heat-loss changes with respect to wind speed.

Did you happen to graph H (heat loss) as a function of v (wind speed)? In general, heat loss is greater when standing in faster winds, but how do fluctuations in wind speed affect heat loss? dH/dt gives us a measure, to quantify how a change in v affects H.

With no wind at all, heat loss is 10.5, but, let wind speed increase by just one meter per second, and heat loss goes from 10.5 up to 339.5 -- that is, H changes by 328 when v changes from 0 to 1.

Increasing v by 1 unit again (from 1 m/s to 2 m/s) results in H increasing by only 135.7 instead of by 328. So, H is not increasing by the same amount, for 1-unit increases in v; we see that subsequent heat losses go up by smaller amounts, for each v increase of 1. In other words, heat loss through each square meter of exposed skin continues to increase with increasing wind speeds, but those increases become less and less dramatic.

Therefore, the rate of heat-loss change decreases as wind speed increases. dH/dv gets smaller as v gets bigger.

These changes in H and v are smooth and continuous, so dH/dv changes smoothly, too. A small fluctuation in v leads to small fluctuations in H and dH/dt. I'm assuming that you have already seen derivative defined as a limit of a difference quotient. My examples above, where v increased by an entire unit, describe average changes in H, over 1-unit jumps in v. Any particular value of the variable dH/dv is a limit of average changes, as fluctuations in the related v go to zero. This limiting value is dH/dv, so a specific value of dH/dv represents the exact rate of change in H at the associated value of v.

Graphically, you can "see" values of dH/dv. They are the slopes of lines tangent to the graph of H at each v. For a particular example, draw a tangent line to the curve of H(v) where v=5. The slope of that line is the value of dH/dv when v=5.

Cheers :)
 
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