# Differentation (Business Calculus)

• Feb 2nd 2008, 05:20 PM
XIII13Thirteen
I can't really figure out what I need to plug in. Here is the problem ...

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

$H=33(10*\sqrt{v}-v+10.45)$

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

a.) Find $\frac{dH}{dv}$and interpret it's meaning in this situation

b.) Find the rates of change of H when v=2 and v=5

Thanks in advance. Word problems tend to confuse me
• Feb 3rd 2008, 12:28 AM
mr fantastic
Quote:

Originally Posted by XIII13Thirteen
I can't really figure out what I need to plug in. Here is the problem ...

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

$H=33(10*\sqrt{v}-v+10.45)$

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

a.) Find $\frac{dH}{dv}$and interpret it's meaning in this situation

b.) Find the rates of change of H when v=2 and v=5

Thanks in advance. Word problems tend to confuse me

(a) Expand and re-write, noting that $\sqrt{v} = v^{1/2}: \,$

$H = 330 v^{1/2} - 33v + (33)(10.45)$

Now use the usual rule for differentiating:

$\frac{dH}{dv} = \frac{1}{2} \times 330 v^{-1/2} - 33 = 165 \times \frac{1}{v^{1/2}} - 33 = \frac{165}{\sqrt{v}} - 33$.

Rate of change of H with respect to v.

(b) Sub v = 2 and v = 5 into the rule for $\frac{dH}{dv}$.