I’m so lost, pls help

For a particle with some position function y(t), we're given its velocity function v(t) = 1 - tan⁻¹(eᵗ ) and we're told that it starts with a position of y = -1 at the start.

(a) Acceleration is the rate of change of velocity, so to get the acceleration function a(t), you have to differentiate v(t) :

a(t) = d/dt v(t)

a(t) = d/dt [1 - tan⁻¹(eᵗ )]

a(t) = - d/dt [tan⁻¹(eᵗ )]

Apply the chain rule: take f(u) = tan⁻¹(u) and u(t) = eᵗ, so that

df/dt = df/du × du/dt

→   a(t) = - 1/(1 + u ²) × eᵗ = - eᵗ / (1 + e²ᵗ )

Then when t = 2, the acceleration is

a (2) = - e² / (1 + e⁴) ≈ -0.1329

(b) The acceleration of the particle at t = 2 is negative, so the particle's speed at this moment is decreasing. (Recall the first derivative test.)

(c) Note that at t = 0, the particle's velocity is v (0) = 1 - tan⁻¹(e⁰) = 1 - π/4 ≈ 0.2416 > 0. This means that the particle is moving upward at the start.

From part (a), we see that the acceleration is always negative, so the particle would be slowing down until it reaches zero velocity, after which point its velocity will become negative (meaning its speed would increase again as it starts to accelerate in the negative direction).

This means the particle's highest position on the y-axis occurs at the same time v(t) becomes 0, so you solve:

1 - tan⁻¹(eᵗ ) = 0

1 = tan⁻¹(eᵗ )

tan(1) = tan(tan⁻¹(eᵗ ))

tan(1) = eᵗ

ln(tan(1)) = ln(eᵗ )

ln(tan(1)) = t ≈ 0.443

(d) Use the fundamental theorem of calculus to find an expression for y(t) :

y(t) = y (0) + ∫₀ᵗ v(s) ds

(where s is just a dummy variable)

Then the particle's position at t = 2 is

y (2) = y (0) + ∫₀² v(s) ds

v(t) unfortunately does not have a nice antiderivative, so you'll have to use a calculator. (I hope this question is given in the calculator portion!) You would find

∫₀² v(s) ds ≈ -0.3607

so that

y (2) ≈ -1 + (-0.3607) ≈ -1.3607

The sign of the particle's velocity at this time tells you in which way the particle is moving. We have

v (2) = 1 - tan⁻¹(e²) ≈ -0.4363

which is negative, so the particle is moving away from the origin. This is consistent with the answer found in part (c):

• before t ≈ 0.443, the particle is moving upward toward the origin;

• at t ≈ 0.443, the particle reaches its highest point at y (2) ≈ -1.3607, then turns around and accelerates back downward and away from the origin

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