Deer in northern Minnesota reproduce according to the linear...

Deer in northern Minnesota reproduce according to the linear differential equation $\frac{d u}{d t}=.27 u$ where $t$ is measured in years. If the initial population is $u(0)=5,000$ and the environment can sustain at most 1,000,000 deer, how long until the deer run out of resources?

Key Concepts

Carrying Capacity

Carrying capacity is a concept in ecology that refers to the maximum number of individuals an environment can sustainably support. It is used to understand the limits imposed by available resources and is a critical parameter when assessing the long-term viability of populations.

Initial Value Problems

Initial value problems combine a differential equation with a given initial condition. This allows for the determination of a unique solution that models the system's behavior starting from a specific initial state.

Logarithmic Transformation

Logarithmic transformations are used to solve equations involving exponential functions. By taking the logarithm on both sides of an equation, one can isolate and solve for a variable in the exponent, such as time in exponential growth scenarios.

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives, representing rates of change. They form the backbone of many modeling techniques, including those used in population dynamics, by providing a framework to describe how a system evolves over time.

Exponential Growth

Exponential growth is a process where a quantity increases by a constant percentage per unit time, leading the quantity to grow at an accelerating rate. In population dynamics, this means that the population size grows faster as it becomes larger, following an exponential function of the form u(t) = u(0)e^(rt).

Transcript

00:01 In this problem of exponential function, we have given that the exponential growth of deer population in massachusetts can be calculated using the formula that is t equals to 50 ,000, this is 50 ,000 multiplied with 1 plus 0 .06 to the power n.

00:23 Where this is the 50 ,000 is the initial population of the deer and 0 .60.

00:33 This is the rate that is 0 .06 is the rate of growth and t is a population after n years and where t is the population after n years and here n is the number of years number of years this is the here modeling formula and now we have to predict the population after four years so that means n is equals to four and now put the value so that should be t is the total population after 4 year that is 50 ,000 multiplied with 1 plus 0 .06 to the power 4 and now evaluated so this would be 50 ,000 multiplied with 1 .06 to the power 4 is equal to this is equal to 63 ,000 so this is equal to 63 ,000 123 .123 .8 848 dears or we can say this is equal to 63 .124 dears.

01:43 So this is the answer or we can say this is approximately this is 63 ,000 dears.

01:57 So this is the answer for part a and now for the part b.

02:03 In the part b, if the initial population was 30 ,000 and growth rate was so if the initial population was 30 ,000 and the growth rate was 0 .12 so the rate was 0 .12 that is the double of that approximately how many days would have been present after 3 years so now we have to find the value after 3 so that means n is equal to 3 now this would be here 30 000 and this would be 1 plus 0 .12 to the power 3 and now when we solve it we get this is approximately equal to 42 2 ,000 deer.

02:46 So this is the answer for part b.

02:50 And now for the part c...

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