Title MATH 23 LE 1 ANSWER KEY.pdf ✅

Math 23 Sample 1st Long Exam First Semester, A.Y. 2025-2026 MBAN 109 Note that fx and fy are continuous within the domain D = {(x, y) : x 3 ≥ y 2 }. In particular, they exist and are continuous at (2, 1). Therefore, f is differentiable at (2, 1). (b) Recall that the formula for the differential of z at a point (x0, y0) is dz := [fx(x0, y0)]dx + [fy(x0, y0)]dy So at the point (x, y), we have dz = [fx(x, y)]dx + [fy(x, y)]dy = 9 2 x 2 p x 3 − y 2 dx + −3y p x 3 − 2y dy and at (2, 1) dz = [fx(2, 1)]dx + [fy(2, 1)]dy = 18√ 7dx − 3 √ 7dy (c) Recall that the local linear approximation of f at (x0, y0) is defined as L(x, y) = f(x0, y0) + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0). In particular, the local linear approximation of f at (2, 1) is L(x, y) = f(2, 1) + fx(2, 1)(x − 2) + fy(2, 1)(y − 1) = 7√ 7 + 18√ 7(x − 2) − 3 √ 7(y − 1). Therefore, L(−0.02, 0.13) ≈ 19.63. Remark: The actual value of f(2.02, 0.98) is f(2.02, 0.98) = (2.023 − 0.982 ) 3 2 ≈ 19.65064127. 7. Suppose ∇g(2, 1 + ln 2) = ⟨−1, 3⟩. Let h(s, t) = g(3s 3 − 2t + 1, st + ln(s 2 + t 2 )). Find ∂h ∂s and ∂h ∂t at (1, 1). If z satisfies z 3 − (s 2 + t 2 )z = 4h(s, t), find ∂z ∂s at (1, 1), assuming h(1, 1) = 0 and z ̸= ± √ 2. Solution: Since h is a function of x and y, and x, y are functions of s and t, then we have by the chain rule, ∂h ∂s = ∂h ∂x ∂x ∂s + ∂h ∂y ∂y ∂s and ∂h ∂t = ∂h ∂x ∂x ∂t + ∂h ∂y ∂y ∂t . We are already given that ∂h ∂x = −1 and ∂h ∂y = 3. So by directly computing, we have ∂h ∂s = −9s 2 + 3 t + 2s s 2 + t 2 = −9s 2 + 3t + 6s s 2 + t 2 ∂h ∂t = 2 + 3 s + 2t s 2 + t 2 = 2 + 3s + 6t s 2 + t 2 Page 4