The size of an animal population at time t is given by $\mathrm { N } ( \mathrm { t } ) = 50,000 + 200 \sin \frac { 7 \pi t } { 48 }$ , where t is the number of months from June 1, 1980. At what rate is the animal population changing on June 1, 1981?
Enter your answer as a quotient reduced form $\frac { a } { b }$ with π in the numerator (no units).

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The Answer of The size of an animal population at time t is given by $\mathrm { N } ( \mathrm { t } ) = 50,000 + 200 \sin \frac { 7 \pi t } { 48 }$ , where t is the number of months from June 1, 1980. At what rate is the animal population changing on June 1, 1981?
Enter your answer as a quotient reduced form $\frac { a } { b }$ with π in the numerator (no units).

The Size of an Animal Population at Time T Is N(t)=50,000+200sin⁡7πt48\mathrm { N } ( \mathrm { t } ) = 50,000 + 200 \sin \frac { 7 \pi t } { 48 }N(t)=50,000+200sin487πt​

The Size of an Animal Population at Time T Is $\mathrm { N } ( \mathrm { t } ) = 50,000 + 200 \sin \frac { 7 \pi t } { 48 }$