hello I need help with the graph in a software of these exercises 1) The force on (x, y) is F* = (2x + y) i + (x + 2y) j. Find the work done by F along the following curves that go from the point (0,0) to (1,3) a) Line connecting the points (0,0) and (1,3) b) Arc of parable joining the points (0,0) and (1,3) The answer for this exercise is: F(x,y)= (2x + y) + (x+24) I hine joining (0,0) & (1,3) y-o 3-o 2-0 1-0 y = 32 dy =3dx dai dy do 11 dw = (2,4) de (2x+ =12y) + (x+24) 5)-( dxt + dys) 2+ (2x+y)dx + (x+2y) dly y) + (2x+3x)dx + ( a +62) (3 da) seda + 21xdX 26x da 11 ( Johor work de 0 (26nda (2625 13 REDMI NOTE 8 PRO AI QUAD CAMERA 2) Problem The force at a point (x, y, z) in three dimensions is given by F(x, y, z) = yi + zj + xk. Find the work done by F(x, y, z) along the twisted cubic x = t, y t?, z = t3 from (0, 0, 0) to (2, 4, 8). The answer to this exercise it's in a book, what I need is the graph in any software Consider, x = 1, y = 1 and 2 = 1 13 We differentiate both sides with respect to t. This gives us dx dy dz 1, = 2t and = 312 dt dt dt dx = dt, dy = 2tdt and dz = 31% dt Consider, F(x, y, k)= yi + zj + xk Now, [F dr = [(vi + 3 + xik). dr ydx + zdy + xdz Now, we substitute the value of x = 1, y =1, z =t, dx = dt, dy = 2tdt and dz = 31dt S ydx + zdy + xdz inc This gives us S 2tdt +1 (2tdt) +1 (31dt) Now, integrating it from 0 to 2 will give us ide ++ (2tdt)+1(31dt) = f(+* +26*+31)dt + 3 = 412 15 412 fF.dr = 15 Therefore, please help me with the graph in any software, there are two exercises, please, I don't have a computer at the moment. hello I need help with the graph in a software of these exercises 1) The force on (x, y) is F* = (2x + y) i + (x + 2y) j. Find the work done by F along the following curves that go from the point (0,0) to (1,3) a) Line connecting the points (0,0) and (1,3) b) Arc of parable joining the points (0,0) and (1,3) The answer for this exercise is: F(x,y)= (2x + y) + (x+24) I hine joining (0,0) & (1,3) y-o 3-o 2-0 1-0 y = 32 dy =3dx dai dy do 11 dw = (2,4) de (2x+ =12y) + (x+24) 5)-( dxt + dys) 2+ (2x+y)dx + (x+2y) dly y) + (2x+3x)dx + ( a +62) (3 da) seda + 21xdX 26x da 11 ( Johor work de 0 (26nda (2625 13 REDMI NOTE 8 PRO AI QUAD CAMERA 2) Problem The force at a point (x, y, z) in three dimensions is given by F(x, y, z) = yi + zj + xk. Find the work done by F(x, y, z) along the twisted cubic x = t, y t?, z = t3 from (0, 0, 0) to (2, 4, 8). The answer to this exercise it's in a book, what I need is the graph in any software Consider, x = 1, y = 1 and 2 = 1 13 We differentiate both sides with respect to t. This gives us dx dy dz 1, = 2t and = 312 dt dt dt dx = dt, dy = 2tdt and dz = 31% dt Consider, F(x, y, k)= yi + zj + xk Now, [F dr = [(vi + 3 + xik). dr ydx + zdy + xdz Now, we substitute the value of x = 1, y =1, z =t, dx = dt, dy = 2tdt and dz = 31dt S ydx + zdy + xdz inc This gives us S 2tdt +1 (2tdt) +1 (31dt) Now, integrating it from 0 to 2 will give us ide ++ (2tdt)+1(31dt) = f(+* +26*+31)dt + 3 = 412 15 412 fF.dr = 15 Therefore, please help me with the graph in any software, there are two exercises, please, I don't have a computer at the moment