The position of a particle moving in a straight line is given by s(t) = (e^(-t))(cos(5t)) for t>0, where t is in seconds. If the particle changes direction at time T seconds, then T must satisfy the equation:

cos(5T) = 0
5T = arctan(-1/5)
5e^(-t) sin(5t) = 0
tan(5T) = -1/5
cos(5T) = 5

I know that a change in direction will be marked by a change from positive to negative or vice versa, but I don't understand the equations the question gives me. Could someone please talk me through this process to find the right answer?

you are correct, as far as you go. When it changes from pos to neg, it will be zero.

Note that the particle changes direction, not position. So, its velocity changes sign. The velocity is given by the derivative.

s = e^-t cos5t
s' = e^-t (-cos5t - 5sin5t)

so, since e^-t is always positive, we need

-cos5t - 5sin5t = 0
tan 5t = -1/5

Well, let's work through these equations step by step.

The given equation for the position of the particle is s(t) = e^(-t) * cos(5t).

To determine when the particle changes direction, we need to find when its velocity changes sign. The velocity of the particle is given by the derivative of the position function, which is v(t) = s'(t).

Let's differentiate the position function with respect to t to find the velocity function:
v(t) = d/dt [e^(-t) * cos(5t)]

To differentiate this, we can use the product rule: (uv)' = u'v + uv'.

Taking the derivative step by step, we have:
v(t) = (-e^(-t) * cos(5t)) + (e^(-t) * (-sin(5t)) * 5)

Simplifying further, we get:
v(t) = -e^(-t) * cos(5t) - 5e^(-t) * sin(5t)

Now, for the particle to change direction, the velocity needs to change sign. Therefore, we need to solve the equation v(t) = 0 for t>0.

So, we set -e^(-t) * cos(5t) - 5e^(-t) * sin(5t) = 0.

Now we have the equation: 5e^(-t) * sin(5t) = -e^(-t) * cos(5t).

To simplify, we can divide both sides of the equation by e^(-t).

So we get: 5 * sin(5t) = -cos(5t).

Now rearrange the equation to isolate the trigonometric function:
5 * sin(5t) + cos(5t) = 0.

This equation is equivalent to: tan(5t) = -1/5.

Now, we can solve for t by taking the inverse tangent of both sides: arctan(tan(5t)) = arctan(-1/5).

Hence, we have: 5t = arctan(-1/5).

Now divide both sides by 5 to solve for t: t = (1/5) * arctan(-1/5).

So, the correct equation from the given options is: 5t = arctan(-1/5).

I hope this explanation helps you understand the process! Let me know if you have any further questions.

To find the point at which the particle changes direction, we need to determine the value of T that satisfies the equation. Let's go through each equation and see which one is correct.

1. cos(5T) = 0:
This equation represents when the cosine of 5T equals zero. The cosine function equals zero at specific angles, such as π/2, 3π/2, 5π/2, etc. So, we solve for T by setting 5T to one of these angles:
5T = π/2, 3π/2, 5π/2, ...
T = π/10, 3π/10, 5π/10, ...
However, none of these solutions satisfy the condition T > 0, so this equation is not correct.

2. 5T = arctan(-1/5):
This equation gives us the angle whose tangent is -1/5. Taking the arctan of -1/5 yields -0.1974 radians (approximately). To find T, we divide this angle by 5:
T = -0.1974 / 5 = -0.0395 (approximately).
Since T must be greater than 0, this equation is also not correct.

3. 5e^(-t) sin(5t) = 0:
This equation can be rearranged to sin(5t) = 0. The sine function equals zero at specific angles, such as 0, π, 2π, etc. So, we solve for t by setting 5t to one of these angles:
5t = 0, π, 2π, ...
t = 0, π/5, 2π/5, ...
However, this equation does not directly give us the value of T. We need to evaluate the equation s(T) = (e^(-T))(cos(5T)), but it does not match any of the given equations. Therefore, this equation is not correct.

4. tan(5T) = -1/5:
This equation represents the angle whose tangent is -1/5. Taking the arctan of -1/5 yields -0.1974 radians (approximately). To find T, we divide this angle by 5:
T = -0.1974 / 5 = -0.0395 (approximately).
Since T must be greater than 0, this equation is not correct.

5. cos(5T) = 5:
This equation represents when the cosine of 5T equals 5. However, the cosine function only ranges from -1 to 1, so there are no solutions to this equation.

Based on the given options, none of the equations provided correctly represent the point at which the particle changes direction.

Certainly! Let's go step by step.

The given position function is s(t) = e^(-t) * cos(5t), which describes the position of a particle moving in a straight line. To determine when the particle changes direction, we need to find the time T such that the particle transitions from moving in one direction to the opposite direction.

To begin, we look at the equation cos(5T) = 0. When the cosine of an angle equals zero, it means that the angle forms a right angle with the x-axis. In other words, at the time T, the particle reaches a maximum or minimum point in its motion.

To find the value of T that satisfies cos(5T) = 0, we can solve it algebraically. By taking the inverse cosine (arccos) of both sides, we get 5T = arccos(0). Since the cosine of 0 is 1, this equation simplifies to 5T = π/2. Divide both sides by 5 to solve for T, giving us T = π/10.

However, it's important to note that there are multiple possible solutions for T since cosine is a periodic function. So, T = π/10 is only one possible time at which the particle changes direction.

Now let's examine the other equations given:

1. The equation 5T = arctan(-1/5) involves the arctan function, which gives us the angle whose tangent is a given value. This equation represents a different angle, not related to our situation of changing direction.

2. The equation 5e^(-t) sin(5t) = 0 involves the sine function, but it does not directly relate to finding the time T at which the particle changes its direction. This equation might be unrelated to the problem or have another purpose.

3. The equation tan(5T) = -1/5 involves the tangent function. While it seems similar to the first equation cos(5T) = 0, it represents a different angle that does not correspond to the particle's change in direction.

4. The equation cos(5T) = 5 appears to be a typographical error. It does not provide a meaningful relationship in this context.

Thus, our focus should be on the equation cos(5T) = 0, and the solution is T = π/10.