a particle moves on the x-axis in such a way that its position at time t is given by x(t)=3t^5-25^3+60t. for what values of t is the particle moving to the left.
a.-2<t<1 only
b.-2<t<-1 and 1<t<2
c.1<t<2 only
d.-1<t<1 and t>2
e.t<-2,-1<t<1, and t>2
Assuming there is a typo and the original function is:
x(t)=3t^5-25t^3+60t
The derivative is:
x'(t)=15*t^4-75*t^2+60
Factor x'(t) and solve for the zeroes of x'(t)=0.
Determine on which interval(s) x'(t) is negative, which means that the particle is moving "backwards", or to the left.
Hint:
There are two such intervals, and two of the zeroes of x'(t) are t=-1 and +1.
Well, let's analyze the situation here. The particle's movement to the left means that its velocity has a negative value. To find the particle's velocity, we need to take the derivative of its position function, x(t).
Taking the derivative of x(t) = 3t^5 - 25t^3 + 60t, we get:
x'(t) = 15t^4 - 75t^2 + 60
Now, we need to find when the velocity, x'(t), is less than zero (negative). To do that, we can set it equal to zero and determine the intervals where it changes its sign.
15t^4 - 75t^2 + 60 < 0
Factoring it out, we get:
15(t^4 - 5t^2 + 4) < 0
Further simplifying:
15(t^2 - 4)(t^2 - 1) < 0
Now, let's check the intervals between the critical points -2, -1, 1, and 2.
For t < -2:
All factors are negative, so the inequality is true in this interval.
For -2 < t < -1:
Only (t^2 - 4) is positive, so the inequality is false in this interval.
For -1 < t < 1:
Both factors (t^2 - 4) and (t^2 - 1) are positive, so the inequality is true in this interval.
For 1 < t < 2:
Only (t^2 - 1) is positive, so the inequality is false in this interval.
For t > 2:
All factors are positive, so the inequality is true in this interval.
Combining the intervals with a true inequality, we get:
t < -2, -1 < t < 1, and t > 2.
So, the answer is option e: t < -2, -1 < t < 1, and t > 2. Now, let me put my clown nose back on! 🤡
To determine the values of t for which the particle is moving to the left, we need to determine when the velocity of the particle is negative. The velocity of the particle is given by the derivative of the position function x(t):
v(t) = x'(t) = d/dt (3t^5 - 25t^3 + 60t)
To find the values of t for which the particle is moving to the left, we need to find the values of t for which v(t) is negative.
Let's find the derivative of x(t) to find v(t):
v(t) = d/dt (3t^5 - 25t^3 + 60t)
= 15t^4 - 75t^2 + 60
Now we can find the values of t for which v(t) is negative:
15t^4 - 75t^2 + 60 < 0
To solve this inequality, we need to factor the quadratic part:
15(t^2 - 5)(t^2 - 2) < 0
Now we can determine the values of t for which this inequality holds true by examining the signs of each factor:
1. For t^2 - 5 < 0, we have t^2 < 5. Taking the square root of both sides, we get t < ±√5. Since we are looking for negative values of t, this simplifies to t < -√5.
2. For t^2 - 2 < 0, we have t^2 < 2. Taking the square root of both sides, we get t < ±√2. Since we are looking for negative values of t, this simplifies to t < -√2.
Therefore, the values of t for which the particle is moving to the left are:
t < -√5 and t < -√2
In interval notation, this can be written as:
(-∞, -√5) U (-∞, -√2)
So the correct option is:
e. t < -2, -1 < t < 1, and t > 2
To determine when the particle is moving to the left, we need to find when its velocity (the derivative of the position function) is negative.
First, let's find the velocity function. The velocity function is given by taking the derivative of the position function with respect to time:
v(t) = d/dt (3t^5 - 25t^3 + 60t)
Differentiating each term separately using the power rule, we get:
v(t) = 15t^4 - 75t^2 + 60
Now we need to find when the velocity function is negative, so we solve the inequality:
v(t) < 0
15t^4 - 75t^2 + 60 < 0
Now, we can factor the quadratic term to make it easier to solve:
15(t^2 - 5)(t^2 - 2) < 0
To determine the sign of the expression, we can set each term equal to zero and find the sign of each interval:
t^2 - 5 = 0 gives t = ±√5
t^2 - 2 = 0 gives t = ±√2
Creating a number line with the critical points at -√5, -√2, √2, and √5, we can test intervals between these points to find the sign of the expression:
When t < -√5:
We choose t = -6 for simplicity. Substituting into the expression, we get:
15(-6)^4 - 75(-6)^2 + 60 = 15 * 1296 + 2700 + 60 = 19980 > 0
So, the expression is positive in this interval.
When -√5 < t < -√2:
We choose t = -3 for simplicity. Substituting into the expression, we get:
15(-3)^4 - 75(-3)^2 + 60 = 15 * 81 - 675 + 60 = 405 - 675 + 60 = -210 < 0
So, the expression is negative in this interval.
When -√2 < t < √2:
We choose t = 0 for simplicity. Substituting into the expression, we get:
15(0)^4 - 75(0)^2 + 60 = 0 + 0 + 60 = 60 > 0
So, the expression is positive in this interval.
When √2 < t < √5:
We choose t = 3 for simplicity. Substituting into the expression, we get:
15(3)^4 - 75(3)^2 + 60 = 15 * 81 - 675 + 60 = 405 - 675 + 60 = -210 < 0
So, the expression is negative in this interval.
When t > √5:
We choose t = 6 for simplicity. Substituting into the expression, we get:
15(6)^4 - 75(6)^2 + 60 = 15 * 1296 + 2700 + 60 = 19980 > 0
So, the expression is positive in this interval.
Based on the sign analysis, we can conclude that the expression is negative when -√5 < t < -√2 and √2 < t < √5. Since those are the intervals when the velocity is negative, the particle is moving to the left.
Therefore, the correct answer is B. -2 < t < -1 and 1 < t < 2.