A ladybug could sting a beetle if the beetle is at most 20 cm away from the bug. One ladybug was 120 cm east of a clint stone, and moving west at 2cm/second, while a beetle 84 cm north of the same stone and moving south at 1 cm/sec.
a) Draw a chart showing the position of the two bugs at time (t hours).
b) When are they closest?
c) Will the beetle get stung?
at time t seconds, the distance z between the bugs is
z^2 = (120-2t)^2 + (84-t)^2
z dz/dt = -2(120-2t) - (84-t) = 5t-324
so dz/dt = 0 at t=64.8
so find z(64.8) and see whether it is greater than 20 cm
Did you make a diagram?
Did you have the clint stone as the origin O?
After t seconds let the southbound beetle be at P
let the ladybug be at R.
You have a right-angled triangle POR
RO = 84 - t cm
PO = 120 - 2t cm
PR^2 = (84-t)^2 + (120 - 2t)^2
2 PR d(PR)/dt = 2(84-t)(-1) + 2(120-2t)(-2)
= 0 for a max/min of PR
2(84-t)(-1) = 2(120-2t)(2)
(84-t)(-1) = (120-2t)(2)
-84 + t = 240 - 4t
5t = 324
t = 64.8 seconds
Plug that into PR^2 = (84-t)^2 + (120 - 2t)^2
and find PR
Is PR ≤ 20 ??
To answer these questions, we can create a chart showing the position of the ladybug and beetle at different times.
a) Position chart:
Time (t) | Ladybug's position (east of the stone) | Beetle's position (north of the stone)
--------------------------------------------------------
0 hours | 120 cm | 84 cm
1 hour | 118 cm | 83 cm
2 hours | 116 cm | 82 cm
... and so on
We can calculate the positions by using the given information that the ladybug is moving west at 2 cm/sec and the beetle is moving south at 1 cm/sec. So, every hour that passes, the ladybug's position decreases by 2 cm, and the beetle's position decreases by 1 cm.
b) To determine when they are closest, we need to calculate the distance between them at different times and find the minimum distance.
Distance between ladybug and beetle:
In a right-angled triangle, the distance (d) between two points (x1, y1) and (x2, y2) can be calculated using the Pythagorean theorem: d^2 = (x2 - x1)^2 + (y2 - y1)^2.
Applying this to our scenario:
At t hours, the ladybug's position is (120 - 2t) cm to the west of the stone, and the beetle's position is (84 - t) cm to the north of the stone. Therefore, the distance between them at time t is:
distance = sqrt((120 - 2t)^2 + (84 - t)^2)
We can calculate this distance for different t values to find when they are closest.
c) To determine if the beetle will get stung, we need to check if the distance between them is less than or equal to 20 cm. If the distance is less than or equal to 20 cm, the beetle will get stung. Otherwise, it will not get stung.