Question

Two athletes run a race. The first athlete starts 50 meters north of a pole and begins running towards it at 5 m/s. At the same time, the second athlete starts at the pole and begins running east at 4 m/s.

Let A (t) be the area formed by Kevin, Mary Joy, and the pole.

As a function of t, A(t)= l l

The greatest area of the triangle occurs at t= l l at which point the area
will be l l square meters.​

Answer 1

To find the area of the triangle formed by the two athletes and the pole at any given time t, we need to first find the position of each athlete at that time.

Let's assume that the first athlete starts at the origin (0,0) and runs towards the pole at (0,50) at a speed of 5 m/s. After time t, the first athlete will be at position (0, 5t+50).

The second athlete starts at the pole (0,50) and runs east at a speed of 4 m/s. After time t, the second athlete will be at position (4t, 50).

The area of a triangle can be calculated as half the product of its base and height. In this case, the base of the triangle is the distance between the two athletes, which can be calculated using the Pythagorean theorem:

distance = sqrt((0 - 4t)^2 + (5t)^2) = sqrt(16t^2 + 25t^2) = sqrt(41t^2)

The height of the triangle is the distance between the pole and the line connecting the two athletes. To find this distance, we need to calculate the equation of the line connecting the two athletes.

The slope of the line is (50 - 5t)/(0 - 4t) = (5t - 50)/(4t), and the y-intercept is 50. Therefore, the equation of the line is:

y = (5t - 50)/(4t) x + 50

To find the distance between the pole and the line, we need to find the perpendicular distance from the pole to the line. This can be calculated using the formula:

distance = |(ax + by + c)/sqrt(a^2 + b^2)|

where a, b, and c are the coefficients of the line equation in the form ax + by + c = 0, and x and y are the coordinates of the pole.

In this case, a = (5t - 50)/(4t), b = -1, c = 50. Plugging these values into the formula, we get:

distance = |(5t - 50)/(4t) * 0 - 1 * 50 + 50| / sqrt((5t - 50)/(4t)^2 + 1^2)

= 50 / sqrt(25/16 + 1)

= 50 / sqrt(41)

Therefore, the area of the triangle at time t is:

A(t) = 1/2 * sqrt(41t^2) * 50 / sqrt(41)

= 25t

To find the greatest area of the triangle, we need to find the value of t that maximizes the function A(t). Taking the derivative of A(t) with respect to t, we get:

dA/dt = 25

This derivative is a constant, which means that A(t) is a linear function of t and has no maximum or minimum values. Therefore, the greatest area of the triangle is infinite and occurs for any value of t.