Solve the problem.
-Under standard conditions, molecules of a gas collide billions of times per second. If each molecule has diameter average distance between collisions is given by
$$L = \frac { 1 } { \sqrt { 2 } \pi t ^ { 2 } n }$$
where $n$, the volume density of the gas, is a constant. Find $\frac { \mathrm { dL } } { \mathrm { dt } }$.
A) $\frac { \mathrm { dL } } { \mathrm { dt } } = \frac { 1 } { 2 \sqrt { 2 } \pi \mathrm { t } ^ { 3 } \mathrm { n } }$
B) $\frac { d L } { d t } = - \frac { 2 } { \sqrt { 2 } \pi t ^ { 3 } n }$
C) $\frac { \mathrm { dL } } { \mathrm { dt } } = \frac { 2 } { \sqrt { 2 } \pi \mathrm { t } ^ { 3 } n }$
D) $\frac { d L } { d t } = - \frac { 1 } { \sqrt { 2 } \pi t n }$ he

Question

Quizplus · Accepted Answer

To find $\frac { \mathrm { dL } } { \mathrm { dt } }$, we differentiate the given equation $L = \frac { 1 } { \sqrt { 2 } \pi t ^ { 2 } n }$ with respect to $t$. Using the chain rule, we get $\frac { \mathrm { dL } } { \mathrm { dt } } = - \frac { 2 } { \sqrt { 2 } \pi t ^ { 3 } n }$, which matches option B.

Solve the Problem L=12πt2nL = \frac { 1 } { \sqrt { 2 } \pi t ^ { 2 } n }L=2​πt2n1​

Solve the Problem $L = \frac { 1 } { \sqrt { 2 } \pi t ^ { 2 } n }$