(Conservative forces) Conservative forces have a very special property that we will explore in this problem. Consider a function U: R² → R which we will call the potential. The conservative force F associated to the potential U is given by (a) Consider the following potential F= -vu. 1 U(x, y) = x² + y²+1 Compute the force F associated to U and sketch the vector field associated to F. (You should produce something like Figure 2.18 from the lecture notes.) (b) Consider two curves C₁ and C₂ with injective parametrizations Y₁: (0,2л) → R² and Y2 (0,2π) R² given by 3 Y₁(t) = (0, Y2(t) = sin(t), 2π 2/7). Sketch the curves C₁ and C2. (c) Compute the curve integrals fc, F. ds and fc, F.ds. Do you notice anything? (d) If you've done things correctly, you should have found Sc, F. ds = Sc₂ F. ds: the answer is independent of the shape of the curve! In fact, you may have noticed that F. ds = F ds (u(0, 1) — U(0,0)) . C₁
(Conservative forces) Conservative forces have a very special property that we will explore in this problem. Consider a function U: R² → R which we will call the potential. The conservative force F associated to the potential U is given by (a) Consider the following potential F= -vu. 1 U(x, y) = x² + y²+1 Compute the force F associated to U and sketch the vector field associated to F. (You should produce something like Figure 2.18 from the lecture notes.) (b) Consider two curves C₁ and C₂ with injective parametrizations Y₁: (0,2л) → R² and Y2 (0,2π) R² given by 3 Y₁(t) = (0, Y2(t) = sin(t), 2π 2/7). Sketch the curves C₁ and C2. (c) Compute the curve integrals fc, F. ds and fc, F.ds. Do you notice anything? (d) If you've done things correctly, you should have found Sc, F. ds = Sc₂ F. ds: the answer is independent of the shape of the curve! In fact, you may have noticed that F. ds = F ds (u(0, 1) — U(0,0)) . C₁