Provide an appropriate response.
-Under standard conditions, molecules of a gas collide billions of times per second. If each molecule has diameteı average distance between collisions is given by
$\mathrm { L } = \frac { 1 } { \sqrt { 2 } \pi \mathrm { t } ^ { 2 } \mathrm { n } }$ where $n$, the volume density of the gas, is a constant. Find $\frac { d ^ { 2 } L } { d t ^ { 2 } }$.
A) $\frac { \mathrm { d } ^ { 2 } \mathrm {~L} } { d t ^ { 2 } } = - \frac { 2 } { \sqrt { 2 } \pi t ^ { 3 } n }$
B) $\frac { d ^ { 2 } L } { d t ^ { 2 } } = \frac { 6 } { \sqrt { 2 } \pi t ^ { 4 } n }$
C) $\frac { \mathrm { d } ^ { 2 } \mathrm {~L} } { \mathrm { dt } ^ { 2 } } = - \frac { 2 } { \sqrt { 2 } \pi \mathrm { t } ^ { 2 } \mathrm { n } }$
D) $\frac { \mathrm { d } ^ { 2 } \mathrm {~L} } { \mathrm { dt } ^ { 2 } } = - \frac { 6 } { \sqrt { 2 } \pi t ^ { 4 } \mathrm { n } }$

Question

Quizplus · Accepted Answer

The second derivative of $L$ with respect to $t$ involves differentiating twice. The first derivative of $L$ with respect to $t$ gives a factor of $-2$ in the numerator due to the power rule, and differentiating again introduces another factor of $-3$, making the overall factor in the numerator $2 \times 3 = 6$, and the $t$ in the denominator becomes $t^4$, resulting in the correct expression being $\frac { 6 } { \sqrt { 2 } \pi t ^ { 4 } n }$.

Provide an Appropriate Response L=12πt2n\mathrm { L } = \frac { 1 } { \sqrt { 2 } \pi \mathrm { t } ^ { 2 } \mathrm { n } }L=2​πt2n1​

Provide an Appropriate Response $\mathrm { L } = \frac { 1 } { \sqrt { 2 } \pi \mathrm { t } ^ { 2 } \mathrm { n } }$