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How quickly is the distance from a baseball to first base changing as the baseball moves along the third baseline

A baseball is moving along the third base line at a constant velocity of 100 feet per second. If a baseball diamond is a 90-foot square, then how quickly is the distance from the ball to first base changing when:

  1. the ball is halfway to third base;
  2. the ball is at third base.

First, we give the general set up for the problem.

    \begin{align*}  x &= \text{ the distance from the ball to first base}. \\  y &= \text{ the distance from the ball to home plate}.  \end{align*}

The following diagram illustrates the situation:

Rendered by QuickLaTeX.com

The quantity we are given then is

    \[ \frac{dy}{dt} = 100 \text{ ft/s}. \]

Further, x^2 = y^2 + 90^2; thus, x = \sqrt{y^2 + 90^2} and so

    \[ \frac{dx}{dy} = \frac{y}{\sqrt{y^2 + 90^2}}. \]

Now, the two requested cases:

  1. If the ball is halfway to third base, this means y = 45 feet (since the diamond is a square with sides of length 90 feet). So we have,

        \[ \frac{dx}{dt} = \frac{dy}{dt} \cdot \frac{dx}{dy} = (100) \left( \frac{45}{\sqrt{45^2 + 90^2}}\right) = \frac{100}{\sqrt{5}} = 20 \sqrt{5} \text{ ft/s}. \]

  2. If the ball is at third base, this means y = 90 (since y is the distance from the ball to home plate), thus we have,

        \[ \frac{dx}{dt} = \frac{dy}{dt} \cdot \frac{dx}{dy} = (100) \left( \frac{90}{\sqrt{90^2 + 90^2}}\right) = \frac{100}{\sqrt{2}} = 50 \sqrt{2} \text{ ft/s}. \]

Compute the velocity of an airplane given the rate of change of its distance to a point on the ground

An airplane is 8 miles above the ground flying at a constant velocity, at a constant altitude. (Assume the earth is flat.) There is a point P on the ground directly under the airplane’s flight path. The distance between P and the airplane is decreasing at a rate of 4 miles per minute when the distance is 10 miles. Find the velocity of the airplane.


Let x be the distance from the point on the ground directly beneath the plane to the point P, and let y be the distance from the plane to P. The following diagram illustrates the situation:

Rendered by QuickLaTeX.com

We are trying to compute \frac{dx}{dt}, the velocity of the airplane. We are given that the distance from the airplane to the point P is changing at a rate of 4 miles per minute when y = 10. Thus, adjusting units to mph, we have

    \[ \frac{dy}{dt} = 240 \text{ mph} \qquad \text{when } y = 10 \text{ miles}. \]

Furthermore, we can compute x^2 in terms of y^2 (by the Pythagorean identity) and then differentiate:

    \[ x^2 = y^2 - 64 \quad \implies \quad \frac{dx}{dy} = \frac{y}{\sqrt{y^2 - 64}}. \]

So, at y = 10 miles we have \frac{dx}{dy} = \frac{5}{3}; thus,

    \[ \frac{dx}{dt} = \left( \frac{dy}{dt} \right) \left( \frac{dx}{dy} \right) = (240) \frac{5}{3} = 400 \text{ mph}. \]

This was computed for y = 10 miles, but since we are given that the airplane is flying at a constant velocity, then this velocity is valid for all y.

Compute the rate of change of volume of a cube given the rate of change of an edge

Given a cube with edges expanding at a rate of 1 centimeter per second, what is the rate of change of the volume of the cube when the length of an edge is:

  1. 5cm.
  2. 10cm.
  3. x cm.

First, we know the volume of a cube is given by

    \[ V = e^3 \]

where e is the length of an edge. Thus, the rate of change of the volume relative to the rate of change of an edge is given by

    \[ \frac{dV}{de} = 3e^2. \]

We use this formula to find the rate of change of the volume for given values of e,

  1. If e = 5 then we have

        \[ \frac{dV}{de} = 3 (5)^2 = 75 \text{ cm}^3 / \text{ sec}. \]

  2. If e = 10 then we have

        \[ \frac{dV}{de} = 3(10)^2 = 300 \text{ cm}^3 / \text{ sec}. \]

  3. If e = x then we have

        \[ \frac{dV}{de} = 3 x^2 \text{ cm}^3 / \text{ sec}. \]