Solution:

Let \(x\) represent the distance from the wall to the foot of the ladder, and let \(y\) represent the height of the top of the ladder on the wall. By the Pythagorean theorem, we have the relationship:

\[x^2 + y^2 = (\sqrt{13})^2 = 13\]

We are given that \(\frac{dx}{dt} = 0.9\) m/s (the foot is moving away from the wall). We need to find \(\frac{dy}{dt}\) when \(x = 2\).

First, we find \(y\) when \(x = 2\):

\[2^2 + y^2 = 13 \implies 4 + y^2 = 13 \implies y^2 = 9 \implies y = \pm 3\]

We take \(y = 3\) since the wall length cannot be negative.

Now, we differentiate both sides of the equation \(x^2 + y^2 = 13\) with respect to time \(t\):

\[\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(13) \implies 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\]

Plugging in the known values \(x = 2\), \(y = 3\), and \(\frac{dx}{dt} = 0.9\):

\[2\cdot 2\cdot 0.9 + 2\cdot 3\cdot \frac{dy}{dt} = 0 \implies 3.6 + 6\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = \frac{-3.6}{6} = -0.6\]

Therefore, the top of the ladder is sliding down the wall at a rate of \(0.6\) m/s. The negative sign indicates that \(y\) is decreasing (the top is moving downward).

Solution:

Let \(x\) represent the horizontal (east-west) distance traveled by ship A, \(y\) represent the vertical (north-south) distance traveled by ship B, and \(z\) represent the distance between the two ships. These quantities are related by the Pythagorean theorem:

\[z^2 = x^2 + y^2\]

At 2:00 pm, 2 hours have passed since noon. In this time, ship A has traveled \(x = 2 \cdot 40 = 80\) km east, and ship B has traveled \(2 \cdot 35 = 70\) km north from its noon position. Note that the vertical separation at 2:00 pm is \(y = 10 + 2 \cdot 35 = 10 + 70 = 80\) km (the initial 10 km separation plus the 70 km that ship B traveled north).

When \(x = 80\) and \(y = 80\), we can find \(z\):

\[z^2 = 80^2 + 80^2 = 6400 + 6400 = 12800 \implies z = \pm \sqrt{12800} = \pm 80\sqrt{2}\]

We take \(x=80\sqrt{2}\), as distance cannot be negative. Now, we differentiate the equation \(z^2 = x^2 + y^2\) with respect to time:

\[2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}\]

We know that \(\frac{dx}{dt} = 40\) km/h and \(\frac{dy}{dt} = 35\) km/h. At 2:00 pm, \(x = 80\), \(y = 80\), and \(z = 80\sqrt{2}\). Substituting:

\[2\cdot 80\sqrt{2}\cdot \frac{dz}{dt} = 2\cdot 80\cdot 40 + 2\cdot 80\cdot 35\implies \sqrt{2}\cdot \frac{dz}{dt} = 40 + 35\implies \frac{dz}{dt} = \frac{75}{\sqrt{2}}=\frac{75\sqrt{2}}{2}\]

Therefore, the distance between the ships is changing at a rate of \(\frac{75\sqrt{2}}{2} \approx 53.03\) km/h at 2:00 pm.

Solution:

Let \(\theta\) represent the angle of elevation, \(x\) represent the horizontal distance from the observer to the point directly below the kite, and \(y = 200\) feet represent the constant height of the kite. From trigonometry, we have:

\[\cot(\theta) = \frac{x}{y} = \frac{x}{200}\]

Now we differentiate the equation \(\cot(\theta) = \frac{x}{200}\) with respect to time:

\[-\csc^2(\theta)\frac{d\theta}{dt} = \frac{1}{200}\frac{dx}{dt}\]

We are given that \(\frac{dx}{dt} = 5\) ft/s. When \(\theta = \frac{\pi}{6}\), we have that \(\csc\left(\frac{\pi}{6}\right) = 2\), and so \(\csc^2\left(\frac{\pi}{6}\right) = 4\). Substituting:

\[-4\cdot \frac{d\theta}{dt} = \frac{1}{200}\cdot 5\implies \frac{d\theta}{dt} = -\frac{1}{160}\]

Therefore, the angle of elevation is changing at a rate of \(-\frac{1}{160}\) radians per second. The negative sign indicates that the angle is decreasing (as the kite moves horizontally away, the angle of elevation decreases).

Solution:

Let \(x\) represent the distance the player has traveled from first base toward second base, \(y = 90\) feet represent the fixed distance from home plate to first base, and \(z\) represent the distance from home plate to the player's current position. By the Pythagorean theorem:

\[z^2 = x^2 + y^2 = x^2 + 90^2\]

When the player is halfway between first and second base, \(x = 45\) feet. At this point:

\[z^2 = 45^2 + 90^2 = 2025 + 8100 = 10125 \implies z = \pm\sqrt{10125} = \pm45\sqrt{5}\]

We take \(z=45\sqrt{5}\), as distance cannot be negative. Now we differentiate the equation \(z^2 = x^2 + 90^2\) with respect to time:

\[2z\frac{dz}{dt} = 2x\frac{dx}{dt}\]

We are given that \(\frac{dx}{dt} = 12\) ft/s. When \(x = 45\) and \(z = 45\sqrt{5}\):

\[2\cdot 45\sqrt{5}\cdot \frac{dz}{dt} = 2\cdot 45\cdot 12\implies\frac{dz}{dt}=\frac{12}{\sqrt{5}}=\frac{12\sqrt{5}}{5}\]

Therefore, the player's distance from home base is increasing at a rate of \(\frac{12\sqrt{5}}{5} \approx 5.37\) ft/s when the player is halfway between first and second base.