The oxygen consumption of a bird embryo increases from the time the egg is laid through the time the chick hatches. In a typical galliform bird, the oxygen consumption (in milliliters per hour) can be approximated by
$c ( t ) = 0.00279 t ^ { 3 } + 0.133 t ^ { 2 } - 0.893 t + 0.14 , ( 8 \leq t \leq 30 )$
Where t is the time (in days) since the egg was laid. (An egg will typically hatch at around $t = 28$ .)

Find $c ^ { \prime } ( 26 )$ . Round your answer to two decimal places.

A) $c ^ { \prime } ( 26 ) = 3.63$
B) $c ^ { \prime } ( 26 ) = 8.22$
C) $c ^ { \prime } ( 26 ) = 11.68$
D) $c ^ { \prime } ( 26 ) = 0.36$
E) $c ^ { \prime } ( 26 ) = - 2.15$

Question

The oxygen consumption of a bird embryo increases from the time the egg is laid through the time the chick hatches. In a typical galliform bird, the oxygen consumption (in milliliters per hour) can be approximated by 
 $c ( t ) = 0.00279 t ^ { 3 } + 0.133 t ^ { 2 } - 0.893 t + 0.14 , ( 8 \leq t \leq 30 )$  
Where t is the time (in days) since the egg was laid. (An egg will typically hatch at around $t = 28$ .)
 
Find $c ^ { \prime } ( 26 )$ . Round your answer to two decimal places.
 
A) $c ^ { \prime } ( 26 ) = 3.63$
B) $c ^ { \prime } ( 26 ) = 8.22$
C) $c ^ { \prime } ( 26 ) = 11.68$
D) $c ^ { \prime } ( 26 ) = 0.36$
E) $c ^ { \prime } ( 26 ) = - 2.15$

Quizplus · Accepted Answer

To find $c'(26)$, we first differentiate $c(t) = 0.00279t^3 + 0.133t^2 - 0.893t + 0.14$ with respect to $t$, which gives $c'(t) = 0.00837t^2 + 0.266t - 0.893$. Substituting $t = 26$ into $c'(t)$ yields $c'(26) = 0.00837(26)^2 + 0.266(26) - 0.893 = 0.00837(676) + 6.916 - 0.893 = 5.66172 + 6.916 - 0.893 = 11.68472$, which rounds to $11.68$.

The Oxygen Consumption of a Bird Embryo Increases from the Time