An ant is walking along the cartesian plane. It starts at the point A=(−18,8) walks in a straight line to a point on
the x-axis, walks directly to the right for 7 units and then
walks in a straight line to the point B=(9,13). What is
the shortest distance that the ant could have walked?
Let the first point on the x axis be P(x,0),
he then goes to Q(x-7,0) and finally to B(9,13)
So the path is AP + PQ + QB
Distance (D) = √( (x+18)^2 + 64) + 7 + √( (x-2)^2 + 16)
dD/dx = .......
= (x+18)/√(x^2 + 36x + 388) + (x-2)/√(x^2 - 4x + 173) . leaving it up to you to simplify to this point
= 0 for a max/min of D
(x+18)/√(x^2 + 36x + 388) = (2-x)/√(x^2 - 4x + 173)
squaring both sides and simplifying
x^4 + 32x^3 + 353x^2 + 4932x + 56052 =
x^4 + 32x^3 + 248x^2 - 1408x + 1552
PHeeww!
105x^2 + 6340x +54500 = 0
21x^2 + 1268x + 10900 = 0
believe it or not, it factored, (good sign I did not make an error)
(21x+218)(x-50) = 0
x = -218/21 or appr -10.38 which makes sense
or
x = 50 , clearly not a minimum
so the ant should go to P(-218/21, 0) and proceed as instructed in your problem
to find the shortest distance, sub in x = -218/21 into the distance expression above
(I will let you do this, it is just tedious arithmetic)
To find the shortest distance that the ant could have walked, we need to determine the path that minimizes the total distance traveled.
Let's break down the given information step by step:
1. The ant starts at point A, which has coordinates (-18, 8).
2. It walks in a straight line to a point on the x-axis.
3. It then walks directly to the right for 7 units.
4. Finally, it walks in a straight line to point B, which has coordinates (9, 13).
First, let's find the point on the x-axis where the ant changes direction. Since the ant walks in a straight line, the y-coordinate of the point on the x-axis will be the same as the y-coordinate of point A, which is 8. Therefore, the ant changes direction at the point (x, 8).
Next, let's find the distance covered from A to the x-axis. Since the ant moves vertically, the distance is the difference between the y-coordinates of A and the point on the x-axis. In this case, it is 8.
Now, the ant moves horizontally to the right for 7 units, so the distance covered in this step is 7.
Lastly, the ant moves in a straight line from the point on the x-axis to point B. Again, the distance is the difference between the x-coordinates of the two points. In this case, it is 9 - x, where x is the x-coordinate of the point on the x-axis.
To minimize the total distance traveled, we need to find the smallest value of x that satisfies the given conditions.
Let's calculate the three distances traveled:
- Distance from A to the x-axis: 8 units
- Distance from the x-axis to B: |9 - x| units
- Horizontal distance from A to B: 7 units
Now, the total distance traveled is given by the sum of these three distances: 8 + |9 - x| + 7.
To minimize this distance, we need to find the smallest value of x that minimizes the expression 8 + |9 - x| + 7. This can be done by considering two cases:
1. Case 1: 9 - x ≥ 0
In this case, the expression |9 - x| simplifies to 9 - x.
So, the total distance is 8 + (9 - x) + 7 = 24 - x.
2. Case 2: 9 - x < 0
In this case, the expression |9 - x| simplifies to -(9 - x), which is equal to x - 9.
So, the total distance is 8 + (x - 9) + 7 = 6 + x.
We need to find the minimum of the total distances obtained in the two cases.
Case 1: Minimum distance = 24 - x
Case 2: Minimum distance = 6 + x
Since we are looking for the shortest distance, we need to find the minimum value between 24 - x and 6 + x. This occurs when both expressions are equal, which means 24 - x = 6 + x.
Solving this equation, we get x = 9.
Therefore, the shortest distance that the ant could have walked is obtained by substituting x = 9 into either expression.
Case 1: Minimum distance = 24 - 9 = 15 units
Case 2: Minimum distance = 6 + 9 = 15 units
In either case, the shortest distance the ant could have walked is 15 units.