91Ó°ÊÓ

Let's denote the distance from H.M.S. Dreadnaught to Montauk as \(x(t)\) and the distance from U.S.S. Mona Lisa to Montauk as \(y(t)\). We are given that \(x(0) = 40\) miles and \(y(0) = 50\) miles. Since the ships are steaming due north and east, their positions change with time at a constant rate. Therefore, we can express \(x(t)\) and \(y(t)\) as follows: \(x(t) = 40 + 20t\) \(y(t) = 50 + 30t\) Now, we need to find the distance between the ships at any time \(t\). We can do this by using the Pythagorean theorem. Let \(z(t)\) denote the distance between the ships at time \(t\). Then: \(z(t)^2 = x(t)^2 + y(t)^2\) Substitute the expressions for x(t) and y(t) above: \(z(t)^2 = (40 + 20t)^2 + (50 + 30t)^2\)

To find the rate at which the distance is increasing, we need to differentiate both sides of the equation with respect to time. \(\frac{dz^2}{dt^2} = \frac{d}{dt} [(40 + 20t)^2 + (50 + 30t)^2]\) Using the chain rule: \(2z \frac{dz}{dt} = 2(40 + 20t)(20) + 2(50 + 30t)(30)\) Now, we are interested in \(\frac{dz}{dt}\), so we need to divide both sides of the equation by \(2z\): \(\frac{dz}{dt} = \frac{(40 + 20t)(20) + (50 + 30t)(30)}{z}\) To find the rate of increase of the distance between the ships at any given time, we need to substitute the values for \(x\) and \(y\), as well as the value for \(z\). But first, let's simplify the numerator of the expression: \(400 + 800t + 1500t + 900t^2 = 1901t^2 + 2300t + 400\) Now, substitute the values of \(x\) and \(y\) from Step 1: \(\frac{dz}{dt} = \frac{1901t^2 + 2300t + 400}{\sqrt{(40 + 20t)^2 + (50 + 30t)^2}}\) Since both ships are moving at a constant speed in their respective directions, we can plug in any value of \(t\) to find out the rate at which their distance is increasing. Note that the denominator will never be zero because \(x^2 + y^2 > 0\) for any \(t\).