University:
California State University, Northridge
Course:
MATH 351 | Differential Equations
Academic year:

2024
Views:

311

Pages:

8
Author:

Javon Neal

Question 1 Find the first partial derivatives of the function. z = xsin(xy) Answer 1 Let f(x, y) = xsin(xy) To find fₓ (treating y as constant and differentiating with respect to x): fₓ = sin(xy) + xycos(xy) To find fᵧ (treating x as constant and differentiating with respect to y): fᵧ = x²cos(xy) Result: fₓ = sin(xy) + xycos(xy) fᵧ = x²cos(xy) Answer 2 Using the chain rule to find partial derivatives: To find ∂z/∂x: ∂z/∂x = ∂/∂x [xsin(xy)] = sin(xy) + xcos(xy) ∂(xy)/∂x = sin(xy) + xycos(xy) To find ∂z/∂y: ∂z/∂y = ∂/∂y [xsin(xy)] = xcos(xy) ∂(xy)/∂y = xcos(xy) * x = x²cos(xy) Final Result: ∂z/∂x = sin(xy) + xycos(xy) ∂z/∂y = x²cos(xy) Answer 3 To differentiate z = xsin(xy) with respect to x and y: For ∂z/∂x: ∂z/∂x = sin(xy) + xycos(xy) For ∂z/∂y: ∂z/∂y = x²cos(xy) Result: ∂z/∂x = sin(xy) + xycos(xy) ∂z/∂y = x²cos(xy) Answer 4 Using the product rule and chain rule: For ∂z/∂x: ∂z/∂x = sin(xy) + xcos(xy) ∂(xy)/∂x = sin(xy) + xycos(xy) For ∂z/∂y: ∂z/∂y = xcos(xy) ∂(xy)/∂y = x²cos(xy) Final Result: ∂z/∂x = sin(xy) + xycos(xy) ∂z/∂y = x²cos(xy) Question 2 Find the derivative of the function y = eˣxᵉ. Answer Given: y = eˣxᵉ Using product and chain rules: Step 1: We know: d/dx (eˣxᵉ) = eˣ d/dx (xᵉ) + xᵉ d/dx (eˣ) Step 2: d/dx (xᵉ) = e xᵉ⁻¹ and d/dx (eˣ) = eˣ. Thus: d/dx (eˣxᵉ) = eˣ [e xᵉ⁻¹] + xᵉ eˣ Simplify: y'(x) = eˣxᵉ⁻¹ + eˣxᵉ = eˣxᵉ⁻¹ (1 + x) Question 3 A man 6 feet tall walks at a rate of 5 ft/s away from a light that is 15 ft above the ground. (a) When he is 10 feet from the base of the light, at what rate is the tip of his shadow moving? (b) At what rate is the length of his shadow changing? Answer 1 Using similar triangles: 15 / (x + s) = 6 / s Cross-multiply and solve for s: 15s = 6(x + s) 15s = 6x + 6s 9s = 6x s = 2/3 x Differentiate with respect to time: ds/dt = (2/3) dx/dt Given dx/dt = 5 ft/s: ds/dt = (2/3)(5) = 10/3 ft/s. (a) Tip of shadow’s movement: Rate = dx/dt + ds/dt = 5 + 10/3 = 25/3 ft/s. (b) Length of shadow’s rate: ds/dt = 10/3 ft/s. Answer 2 Using Pythagoras and similar triangles, find the rate of change of the shadow and tip. Detailed breakdown results in: (a) Tip of shadow moving at 25/3 ft/s. (b) Length of shadow changing at 10/3 ft/s. Question 4 Find the total differential. z = xcos(y) – ycos(x) Answer The total differential is: dz = ∂z/∂x dx + ∂z/∂y dy ∂z/∂x: ∂z/∂x = ∂/∂x [xcos(y) – ycos(x)] = cos(y) + ysin(x). ∂z/∂y: ∂z/∂y = ∂/∂y [xcos(y) – ycos(x)] = –xsin(y) – cos(x). Substitute into dz: dz = [cos(y) + ysin(x)] dx + [–xsin(y) – cos(x)] dy. Question 5 Find the derivative of the vector function: r(t) = e²ᵗi – j + ln(1 + 3t)k. Answer Differentiate each component: dr(t)/dt = d/dt (e²ᵗ)i + d/dt (–j) + d/dt [ln(1 + 3t)]k = 2e²ᵗi + 0j + (3 / (1 + 3t))k. Result: dr(t)/dt = 2e²ᵗi + (3 / (1 + 3t))k.