At 0 Celsius, the heat loss H
(in kilo calories per square meter per hour)from a person's
body can be modeled by
H=33(10 radical V-V+10.45)
where V is the wind speed (in meters per second).
(a) Find dH/dV and interpret its meaning this situation.
(b) Find the rates of change of H when V = 2 and when V=5

Question

At 0 Celsius, the heat loss H
(in kilo calories per square meter per hour)from a person's
body can be modeled by
H=33(10 radical V-V+10.45)
where V  is the wind speed (in meters per second).
(a) Find dH/dV and interpret its meaning this situation.
(b) Find the rates of change of H when V = 2 and when V=5

anonymous · Answer

$$H=33( 10 ^{ V-V+10.45}) ?$$

anonymous · Answer

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

anonymous · Answer

$${dH \over dV} = 33( {5 \over {\sqrt{v}}} -1)$$

anonymous · Answer

it stands for the steepness of the original graph (rate of change)

anonymous · Answer

for part b simply sub 2 and 5

anonymous · Answer

please explain more

anonymous · Answer

sry.can you tell me that I sub2 and 5 to H or dH/dV??

anonymous · Answer

sub it in the dH/Dv since the question is about the rate of change, not the value