Vector Calculus and Mechanics - Study Notes

Review of Vectors

  • Vector vs Scalar
    • Scalar quantities are described by magnitude only (examples: mass, time, speed, length).
    • Vector quantities require both magnitude and direction (examples: displacement, velocity, acceleration).
  • Applications in physics and engineering rely on distinguishing scalars and vectors for analysis and problem solving.

Vector Calculus

  • Core operations in vector calculus: Differentiation, Integration, Gradient, Divergence, Curl.
  • Differentiation types:
    • Ordinary differentiation: deals with functions of a single variable (e.g., f(x)); finds rate of change with respect to that variable.
    • Partial differentiation: deals with functions of multiple variables (e.g., f(x, y, z)); finds rate of change with respect to one variable while holding others constant.
  • Del operator (Vector differential operator):
    • Denoted by ∇ and defined as
    • \nabla = \hat\mathbf{i}\frac{\partial}{\partial x} + \hat\mathbf{j}\frac{\partial}{\partial y} + \hat\mathbf{k}\frac{\partial}{\partial z}

Gradient

  • For a scalar function φ(x, y, z), the gradient is the vector field
  • ∇ϕ=(∂ϕ∂x,∂ϕ∂y,∂ϕ∂z)=∂ϕ∂xi^+∂ϕ∂yj^+∂ϕ∂zk^\nabla \phi = \left(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y},\frac{\partial \phi}{\partial z}\right) = \frac{\partial \phi}{\partial x}\hat{\mathbf{i}} + \frac{\partial \phi}{\partial y}\hat{\mathbf{j}} + \frac{\partial \phi}{\partial z}\hat{\mathbf{k}}
  • The gradient points in the direction of greatest rate of increase of φ and its magnitude is the rate of that increase.
  • Directional derivatives are related to the gradient:
  • Du^ϕ=∇ϕ⋅u^D_{\hat{u}}\phi = \nabla \phi \cdot \hat{u} where \hat{u} is a unit vector in the desired direction.

Numerical: Gradient and Directional Derivative (practice examples)

  • Example 1: If $F(x,y,z) = 3x^2y - y^3z^2$, find grad F at (1, -2, -1).
    • Compute partial derivatives:
    • ∂F∂x=6xy\frac{\partial F}{\partial x} = 6xy
    • ∂F∂y=3x2−3y2z2\frac{\partial F}{\partial y} = 3x^2 - 3y^2 z^2
    • ∂F∂z=−2y3z\frac{\partial F}{\partial z} = -2y^3 z
    • Evaluate at (1, -2, -1):
    • ∂F∂x=6(1)(−2)=−12\frac{\partial F}{\partial x} = 6(1)(-2) = -12
    • ∂F∂y=3(1)2−3(−2)2(−1)2=3−12=−9\frac{\partial F}{\partial y} = 3(1)^2 - 3(-2)^2(-1)^2 = 3 - 12 = -9
    • ∂F∂z=−2(−2)3(−1)=−2(−8)(−1)=−16\frac{\partial F}{\partial z} = -2(-2)^3(-1) = -2(-8)(-1) = -16
    • Therefore, ∇F(1,−2,−1)=(−12, −9, −16).\nabla F(1,-2,-1) = (-12,\,-9,\,-16).
  • Example 2: Directional derivative in direction \mathbf{d} = \langle 2,3,-2\rangle for $A(x,y,z) = 2x z^4 - x^2 y$ at (2, -2, -1).
    • First, gradient of A:
    • ∂A∂x=2z4−2xy\frac{\partial A}{\partial x} = 2 z^4 - 2xy
    • ∂A∂y=−x2\frac{\partial A}{\partial y} = -x^2
    • ∂A∂z=8xz3\frac{\partial A}{\partial z} = 8 x z^3
    • Evaluate at (2, -2, -1):
    • ∂A∂x=2(1)−2(2)(−2)=2+8=10\frac{\partial A}{\partial x} = 2(1) - 2(2)(-2) = 2 + 8 = 10
    • ∂A∂y=−(2)2=−4\frac{\partial A}{\partial y} = -(2)^2 = -4
    • ∂A∂z=8(2)(−1)3=−16\frac{\partial A}{\partial z} = 8(2)(-1)^3 = -16
    • So, ∇A(2,−2,−1)=(10, −4, −16).\nabla A(2,-2,-1) = (10,\,-4,\,-16).
    • Directional derivative along \hat{u} = \frac{\mathbf{d}}{||\mathbf{d}||} with \mathbf{d} = \langle 2,3,-2\rangle, \ ||\mathbf{d}|| = \sqrt{2^2+3^2+(-2)^2} = \sqrt{17}.
    • Du^A=∇A⋅u^=∇A⋅d∣∣d∣∣=(10,−4,−16)⋅(2,3,−2)17=4017.D_{\hat{u}}A = \nabla A \cdot \hat{u} = \frac{\nabla A \cdot \mathbf{d}}{||\mathbf{d}||} = \frac{(10, -4, -16) \cdot (2,3,-2)}{\sqrt{17}} = \frac{40}{\sqrt{17}}.
  • Summary: grad F at (1,-2,-1) is $(-12,-9,-16)$; grad A at (2,-2,-1) is $(10,-4,-16)$; directional derivative of A in direction (2,3,-2) is $\displaystyle D_{\hat{u}}A = \frac{40}{\sqrt{17}}$.

Divergence

  • Divergence of a vector field v with components $(vx, vy, v_z)$ is
  • ∇⋅v=∂v<em>x∂x+∂v</em>y∂y+∂vz∂z.\nabla \cdot \mathbf{v} = \frac{\partial v<em>x}{\partial x} + \frac{\partial v</em>y}{\partial y} + \frac{\partial v_z}{\partial z}.
  • Interpretation:
    • Positive divergence: source field (net outward flow).
    • Negative divergence: sink field (net inward flow).
    • Zero divergence: solenoidal field (no net outflow/inflow).

Curl

  • Curl of a vector field \mathbf{A} with components $(Ax, Ay, A_z)$ is
  • ∇×A=(∂A<em>z∂y−∂A</em>y∂z, ∂A<em>x∂z−∂A</em>z∂x, ∂A<em>y∂x−∂A</em>x∂y).\nabla \times \mathbf{A} = \left( \frac{\partial A<em>z}{\partial y} - \frac{\partial A</em>y}{\partial z},\ \frac{\partial A<em>x}{\partial z} - \frac{\partial A</em>z}{\partial x},\ \frac{\partial A<em>y}{\partial x} - \frac{\partial A</em>x}{\partial y} \right).
  • Interpretation:
    • Curl measures the rotation of a vector field.
    • Zero curl (irrotational) means the field has no local rotation.

Numerical: Curl and Divergence (example problems)

  • Example 1: For (F(x,y,z) = x z^{3}\,\mathbf{i} - 2x^{2} y z\,\mathbf{j} + 2 y z^{4}\,\mathbf{k}), find curl F at ((1,-1,1)).
    • Let F = (F1, F2, F3) with F1 = x z^3, F2 = -2 x^2 y z, F3 = 2 y z^4.
    • Curl components:
    • ∇×F=(∂F<em>3∂y−∂F</em>2∂z, ∂F<em>1∂z−∂F</em>3∂x, ∂F<em>2∂x−∂F</em>1∂y).\nabla\times F = \left( \frac{\partial F<em>3}{\partial y} - \frac{\partial F</em>2}{\partial z}, \ \frac{\partial F<em>1}{\partial z} - \frac{\partial F</em>3}{\partial x}, \ \frac{\partial F<em>2}{\partial x} - \frac{\partial F</em>1}{\partial y} \right).
    • Compute: \nabla×F = (2 z^4 + 2 x^2 y, 3 x z^2, -4 x y z).
    • Evaluate at (1, -1, 1): (0, 3, 4).
  • Example 2: Evaluate div B where (B = \langle 2x^2 z, -xy^2 z, 3 y z^2 \rangle).
    • Divergence: ∇⋅B=∂∂x(2x2z)+∂∂y(−xy2z)+∂∂z(3yz2)=4xz−2xyz+6yz.\nabla\cdot B = \frac{\partial}{\partial x}(2x^2 z) + \frac{\partial}{\partial y}(-x y^2 z) + \frac{\partial}{\partial z}(3 y z^2) = 4 x z - 2 x y z + 6 y z.
    • Note: If evaluated at (1, -1, 1) this gives 4(1)(1)−2(1)(−1)(1)+6(−1)(1)=4+2−6=0.4(1)(1) - 2(1)(-1)(1) + 6(-1)(1) = 4 + 2 - 6 = 0.

Line Integration

  • Line integral of a vector field along a curve C:
  • ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r}
  • Here, the line integral represents the work done by the field along the path or the flux of the field along the path depending on context.

Mechanics

  • Mechanics is the study of motion and the causes of motion.
  • Classical Mechanics
    • Deals with macroscopic objects and their motions under forces.
  • Quantum Mechanics
    • Deals with microscopic systems and their probabilistic behavior.

Newton's Laws of Motion

  • 1st Law (Law of Inertia): A body at rest stays at rest and a body in motion stays in uniform motion unless acted upon by an external unbalanced force.
  • 2nd Law (Law of Acceleration): The acceleration of a body is in the direction of the applied force and is proportional to the net external force and inversely proportional to the mass: F=ma\mathbf{F} = m \mathbf{a}
  • 3rd Law (Action-Reaction): For every action, there is an equal and opposite reaction.

Applications of Newton's Laws

  • Case 01: Atwood machine with two unequal masses, pulley is light and frictionless.
    • Acceleration magnitude: a=(m<em>1−m</em>2)gm<em>1+m</em>2a = \frac{(m<em>1 - m</em>2) g}{m<em>1 + m</em>2}
    • String tension: T=2m<em>1m</em>2gm<em>1+m</em>2T = \frac{2 m<em>1 m</em>2 g}{m<em>1 + m</em>2}
  • Case 02: One mass moves vertically and the other mass moves on a smooth horizontal surface.
    • Equations yield: a=m<em>1gm</em>1+m2a = \frac{m<em>1 g}{m</em>1 + m_2}
    • Direction: the vertical mass moves downward, the horizontal mass moves to the right (assuming the vertical mass is the one pulling the string).
  • Case 03: A traffic light weighing 122 N hangs from a cable tied to two upper cables that connect to a support.
    • Upper cables make angles of 37.0° and 53.0° with the horizontal.
    • Upper cables have a maximum tension of 100 N (they break if tension > 100 N); the vertical cable is stronger.
    • Determine whether the light remains hanging or a cable breaks.
    • Assumptions: All three cables meet at a common junction with the traffic light; the system is in static equilibrium.
    • Let T1 be tension in the cable at 37°, T2 be tension in the cable at 53°, and T_v the vertical cable tension.
    • Equilibrium equations:
    • Horizontal balance: T<em>1cos⁡37∘=T</em>2cos⁡53∘T<em>1 \cos 37^{\circ} = T</em>2 \cos 53^{\circ}
    • Vertical balance: T<em>v+T</em>1sin⁡37∘+T2sin⁡53∘=122.T<em>v + T</em>1 \sin 37^{\circ} + T_2 \sin 53^{\circ} = 122.
    • Worst-case for upper cables (to test breaking): set T_v = 0 (all weight carried by the two upper cables).
    • Solve for T1 and T2 with the horizontal relation: T<em>1=T</em>2cos⁡53∘cos⁡37∘.T<em>1 = T</em>2 \frac{\cos 53^{\circ}}{\cos 37^{\circ}}.
    • Substitute into vertical balance: T2(cos⁡53∘cos⁡37∘sin⁡37∘+sin⁡53∘)=122.T_2\left(\frac{\cos 53^{\circ}}{\cos 37^{\circ}}\sin 37^{\circ} + \sin 53^{\circ}\right) = 122.
    • Using numerical values: (\sin 37^{\circ} \approx 0.6018), (\cos 37^{\circ} \approx 0.7986), (\sin 53^{\circ} \approx 0.7986), (\cos 53^{\circ} \approx 0.6018).
    • Compute the combined factor: ( (\cos 53^{\circ}/\cos 37^{\circ})\sin 37^{\circ} + \sin 53^{\circ} \approx 0.7536 \times 0.6018 + 0.7986 \approx 1.2526 ).
    • Therefore, T<em>2=1221.2526≈97.4 N.T<em>2 = \frac{122}{1.2526} \approx 97.4 \text{ N}. Then T</em>1=T2cos⁡53∘cos⁡37∘≈97.4×0.7536≈73.4 N.T</em>1 = T_2 \frac{\cos 53^{\circ}}{\cos 37^{\circ}} \approx 97.4 \times 0.7536 \approx 73.4 \text{ N}.
    • Vertical tension: T<em>v=122−(T</em>1sin⁡37∘+T2sin⁡53∘)≈122−(73.4×0.6018+97.4×0.7986)≈0 NT<em>v = 122 - (T</em>1 \sin 37^{\circ} + T_2 \sin 53^{\circ}) \approx 122 - (73.4\times 0.6018 + 97.4\times 0.7986) \approx 0\text{ N} (within rounding).
    • Result: The two upper cables carry about 73.4 N and 97.4 N; both are below the 100 N breaking threshold. The vertical cable bears essentially all the remaining load (≈0 N in this ideal calculation), so the traffic light remains hanging and no upper cable breaks.
    • Practical takeaway: If the weight increases or angles change such that either T1 or T2 exceeds 100 N, a cable would break; with W = 122 N and given angles, neither upper cable breaks.