suppose that the position functions of 2 bugs B1 and B2 in motion along the same line are B1=1/2t^2-t+3 and B2=-1/4t^2+t+1 for t is greater than or equal to zero.
a) prove the bugs will not collide
b)how close do the bugs get to each other?
I would greatly appreciate if you could help me I am very confused and unsure how to solve this problems.
Sure, I can help you with these problems. Let's start with part (a).
a) To prove that the bugs will not collide, we need to show that their positions will never be the same at the same time. In other words, we need to show that there is no value of t for which B1(t) = B2(t).
To do this, we can set up the equation B1(t) = B2(t) and solve for t:
1/2t^2 - t + 3 = -1/4t^2 + t + 1
Multiplying both sides of the equation by 4 to get rid of the fractions:
2t^2 - 4t + 12 = -t^2 + 4t + 4
Combining like terms:
3t^2 - 8t + 8 = 0
This is a quadratic equation. We can use the quadratic formula, t = (-b ± √(b^2 - 4ac))/2a, to solve for t. Here, a = 3, b = -8, and c = 8.
t = (-(-8) ± √((-8)^2 - 4(3)(8)))/(2(3))
t = (8 ± √(64 - 96))/6
t = (8 ± √(-32))/6
Since the discriminant (√(64 - 96)) is negative, the quadratic equation has no real solutions. Therefore, there is no value of t for which B1(t) = B2(t). Hence, the bugs will not collide.
b) To find out how close the bugs get to each other, we need to determine the minimum distance between their positions. This can be found by finding the minimum value of the function that represents the distance between the bugs, d(t) = |B1(t) - B2(t)|.
To find the minimum value, we can differentiate d(t) with respect to t and set it equal to zero:
d'(t) = B1'(t) - B2'(t) = 0
Differentiating B1(t) and B2(t):
B1'(t) = t - 1
B2'(t) = -t + 1
Setting them equal to each other:
t - 1 = -t + 1
Simplifying:
2t = 2
t = 1
So, at t = 1, the bugs are closest to each other.
To find the minimum distance, we can substitute t = 1 into the distance function:
d(1) = |B1(1) - B2(1)|
Calculate B1(1) and B2(1) by substituting t = 1 into the position functions of the bugs:
B1(1) = 1/2(1)^2 - (1) + 3 = 1/2 - 1 + 3 = 1/2 + 2 = 2.5
B2(1) = -1/4(1)^2 + (1) + 1 = -1/4 + 1 + 1 = 3/4 + 1 = 1.75
Therefore, the bugs get as close as 1.75 units to each other at t = 1.