Frog Geometry problem

Question

1. A stone is tossed into a pond at the coordinates (0, 0), creating a circular ripple that increases at a rate of 12 cm/s. A frog is on his lily pad at the point (–36, 48). How far is the frog from the place where the stone entered the water? How many seconds will have passed when the ripple reaches the frog? (4 marks)

Answer 1

Here is a rough sketch of the frog and ripples:

We have to find the distance from the stone at (0,0) and the frog at (-36, 48), using the distance formula:

Distance = √((x₂-x₁)²+(y₂-y₁)²)

= √((-36-0)²+(48-0)²)

= √((-36)²+(48)²)

= √((1296+2304)

= √3600

=60

The frog is 60 cm away.

Use this formula to find the time:

Distance = rate * time

60 = 12t

t = 60/12 = 5

It takes 5 seconds for the ripple to reach the frog.

Answer 2

The fundamental math being exercised in this problem is the two dimensional distance between two points. Subsequently, we can calculate the elapsed time for the ripple to travel that distance because we are given the ripple's constant velocity.

The fundamental mathematical principle here is the use of the Pythagorean theorem. We have a pair of two-dimensional points in space and we want to calculate the distance between them. To do so, we simply square the difference in each dimension, add the squares, and then take the square root of the sum.

For this solution I will use common terms instead of traditional mathematical variables...

distance = square_root((change in horizontal position)^2 + (change in vertical position)^2)

This problem is a bit simpler than others because the initial point is the origin of the two-dimensional plane (0,0) so the change of each position is simply equal to the second ordered pair's values.

distance = square_root(36^2 + 48^2) = 60cm!

To find the time elapsed until the ripple reaches the frog we will divide the distance by the ripple velocity to get our answer!

time = distance / ripple velocity = 60cm / 12cm/s = 5s!