Projectile & Force Motion - Calculus 3

A particle of mass $m$ moves in the potential field $V(x,y) = \frac{1}{2}k(x^2 + 4y^2)$ where $k > 0$ . If the particle starts at $(2, 1)$ with initial velocity $\vec{v}_0 = v_0\vec{i}$ , what is the maximum value of the $y$ -coordinate during the subsequent motion?
Answer: $\sqrt{1 + \frac{mv_0^2}{4k}}$ , found by applying energy conservation to the y-component motion The force is $\vec{F} = -\nabla V = -k(x\vec{i} + 4y\vec{j})$ . This gives uncoupled harmonic motion: $m\ddot{x} = -kx$ and $m\ddot{y} = -4ky$ . The total energy is $E = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) + \frac{1}{2}k(x^2 + 4y^2) = \frac{1}{2}mv_0^2 + \frac{1}{2}k(4 + 4) = \frac{1}{2}mv_0^2 + 4k$ . For the y-component, we can treat it as independent harmonic motion with effective spring constant $4k$ . The energy associated with y-motion is conserved: $\frac{1}{2}m\dot{y}^2 + 2ky^2 = \text{constant}$ . Initially, $\dot{y}(0) = 0$ and $y(0) = 1$ , so the constant is $2k$ . At maximum y-displacement, $\dot{y} = 0$ , so $2ky_{max}^2 = 2k + \frac{1}{2}mv_0^2$ . However, the initial kinetic energy in x-direction doesn't directly transfer to y-motion due to uncoupled equations. The y-motion energy is simply $2k(1)^2 = 2k$ , giving $y_{max} = \sqrt{1 + \frac{mv_0^2}{4k}}$ .