Detailed Answer

We know \( y = 5e^{x^4} \) so \( \frac{dy}{dx} = 20x^3e^{x^4} = 20x^3y \). (Chain rule used)

Detailed Answer

a) If \( P=ce^{\frac{t}{2}} \), then \( \frac{dP}{dt} = \frac{c}{2}e^{\frac{t}{2}} = \frac{P}{2} \).

b) The population of alligators cannot be negative.

c) We know when \( t=0 \), \( P=5 \), so plugging into the equation in (a), we find \( c=5 \).

Detailed Answer

a) \( y = \int 6x^2 \, dx = 2x^3 + C \)

b) \( y = \int \frac{1}{2x+6} \, dx = \frac{1}{2}\ln{(2x+6)} + C \)

c) \( y = \int \frac{p}{\sqrt{36-p^2}} \, dp \)
We can do a u-substitution here, letting \( u = 36-p^2 \), so \( \frac{du}{dp}=-2p \), and substituting these in:

\begin{align*} y &= \int -\frac{1}{2}\frac{1}{\sqrt{u}} \, du\\ y &= -\frac{1}{2}\left(2\sqrt{u} + C\right)\\ y &= -\sqrt{36-p^2}+C \end{align*}

d) This is a separable differential equation:

\begin{align*} \frac{dy}{y-1} &= \frac{dx}{x+5} \\ \int \frac{dy}{y-1} &= \int \frac{dx}{x+5} \\ \ln{|y-1|} &= \ln{|x+5|} + C\\ \ln{|y-1|} - \ln{|x+5|} &= C \\ \ln{\left|\frac{y-1}{x+5}\right|} &= C \\ \frac{y-1}{x+5} &= e^C \\ y &= e^C(x+5) + 1 \end{align*}

Note: It is common practice to replace \( e^C \) with another constant, \( A \), for aesthetics.

Detailed Answer

We are told \( \frac{dA}{dt} = k\sqrt[3]{A} \). From here, we can solve the differential equation to get the desired function:

\begin{align*} \frac{1}{\sqrt[3]{A}} dA &= k \, dt \\ \int \frac{dA}{\sqrt[3]{A}} &= \int k \, dt \\ \frac{3}{2} \sqrt[3]{A^2} &= kt + C \end{align*}

We were told that when \( t=0 \), the area was \( 2\sqrt{2} \):

\begin{align*} \frac{3}{2} \sqrt[3]{(2\sqrt{2})^2} &= k(0) + C \\ C &= 3 \end{align*}

When \( t=4 \), we know the area doubled, so it became \( 4\sqrt{2} \):

\begin{align*} \frac{3}{2} \sqrt[3]{(4\sqrt{2})^2} &= k(4) + 3 \\ k &= \frac{-3 + 3\sqrt[3]{4}}{4} \end{align*}

So finally:

\begin{align*} \frac{3}{2} \sqrt[3]{A^2} &= \left(\frac{-3 + 3\sqrt[3]{4}}{4}\right)t + 3 \\ \frac{1}{2} \sqrt[3]{A^2} &= \left(\frac{-1 + 1\sqrt[3]{4}}{4}\right)t + 1 \\ A(t) &= \sqrt{\left( \left(\frac{-1 + 1\sqrt[3]{4}}{2}\right)t + 2 \right)^3} \end{align*}

Detailed Answer

The easiest way to solve this is by creating a table:

\[ x_{n+1} = x_n + 0.25 \]

\[ y_{n+1} = y_n + 0.25(\frac{x_n^2+y_n^2}{2x_n^2}) \]

\[ Error = \frac{Actual\:value - Expected\:value}{Expected\:value}\cdot 100 = \frac{0.641-0.515}{0.515} \cdot 100 = 24.5\% \]

Detailed Answer

a) (i) When \( t=0 \), we have: \(\left(\begin{array}{c} x \\ y \end{array}\right) = \left(\begin{array}{c} 1 \\ 1 \end{array}\right)\) , therefore: \(\mathbf{\dot{x}} = \left(\begin{array}{c} 2 \\ 2 \end{array}\right)\) .

a) (ii) The general solution to a system of this form is: \(\mathbf{x} = Ae^{2t}\left(\begin{array}{c} 1 \\ 1 \end{array}\right) + Be^{3t}\left(\begin{array}{c} 1 \\ 2 \end{array}\right)\) . We know that when \( t=0 \), \( \mathbf{x} = \left(\begin{array}{c} 1 \\ 1 \end{array}\right) \), thus: \(\left(\begin{array}{c} 1 \\ 1 \end{array}\right) = A\left(\begin{array}{c} 1 \\ 1 \end{array}\right) + B\left(\begin{array}{c} 1 \\ 2 \end{array}\right)\) .

Therefore, we can write: \(\left(\begin{array}{c} 1 \\ 1 \end{array}\right) = \left(\begin{array}{cc} 1 & 1\\ 1 & 2 \end{array}\right) \left(\begin{array}{c} A \\ B \end{array}\right)\) .

Solving for \( \left(\begin{array}{c} A \\ B \end{array}\right) \) \( = \left(\begin{array}{cc} 1 & 1\\ 1 & 2 \end{array}\right)^{-1} \left(\begin{array}{c} 1 \\ 1 \end{array}\right) = \left(\begin{array}{c} 1 \\ 0 \end{array}\right)\) , so the particular solution is: \(\mathbf{x} = e^{2t}\left(\begin{array}{c} 1 \\ 1 \end{array}\right)\) .

b) The system will asymptotically approach the line: \(k\left(\begin{array}{c} 1 \\ 1 \end{array}\right)\) , as that is the only term in the particular solution.

Detailed Answer

a) If we let \( y = \frac{dx}{dt} \), then we have: \( \frac{dy}{dt} = \frac{d^2x}{dt^2} \). Thus, the system can be expressed as: \[ \begin{cases} \frac{dx}{dt} = y \\ \frac{dy}{dt} = -\frac{1}{25}y \end{cases} \]

b) We can solve the second equation as follows: \begin{align*} \frac{dy}{dt} &= -\frac{1}{25}y \\ \frac{1}{y}dy &= -\frac{1}{25}dt \\ \int \frac{1}{y}dy &= \int -\frac{1}{25}dt \\ \ln{|y|} &= -\frac{1}{25}t + C\\ y(t) &= e^C \cdot e^{-\frac{1}{25}t} = A \cdot e^{-\frac{1}{25}t} \end{align*}

c) Now we substitute \( y \) back into the equation for \( x \): \begin{align*} \frac{dx}{dt} &= y = A \cdot e^{-\frac{1}{25}t} \\ x(t) &= -25A \cdot e^{-\frac{1}{25}t} + C \end{align*}

Detailed Answer

a) If we set \( y = \frac{dx}{dt} \), then we have: \( \frac{dy}{dt} = \frac{d^2x}{dt^2} \). Thus, the system can be expressed as: \[ \begin{cases} \frac{dx}{dt} = y \\ \frac{dy}{dt} = 9.8 - 0.5y^2 \end{cases} \]

b) At the start, the ball is stationary. Given the initial conditions \( t_0=0 \), \( x_0=0 \), \( y_0=0 \), we need to make 5 steps over 2 seconds, so each step will be: \( h = 0.4 \) seconds. The updates for \( t \), \( x \), and \( y \) are given by: \begin{align*} t_i &= t_{i-1} + 0.4 \\ x_i &= x_{i-1} + 0.4y_{i-1} \\ y_i &= y_{i-1} + 0.4(9.8 - 0.5y_{i-1}^2) \end{align*} The computed values are shown in the table below:

From this table, we can see that \( x_5 = 6.983 \), meaning in the first 5 seconds, it fell approximately 6.983 meters.

c) To find the terminal velocity, we look for the value \( y_n \) approaches as \( n \) goes to infinity. Continuing our table from part (b):

We can see \( y_n \) is approaching 4.43, hence that is our terminal speed.

d) When we reach the terminal speed, the acceleration of the ball becomes 0. Meaning, \( \frac{d^2x}{dt^2} = 0 \), and so,

\[ 0 = 9.81 - 0.5\left(\frac{dx}{dt}\right)^2 \]

\[ \frac{dx}{dt} = \sqrt{\frac{9.8}{0.5}} \approx 4.427\]

This is very close to the solution in part (c).