\f\fAnswer 1: The gradient of the temperature function is T x, y Tx x, y i Ty x, y j 1 1 3 y 16 x i 3x 4 y j 3 3 = To find the critical point, solve Tx x, y 0 and Ty x, y 0 1 1 3 y 16 x 0 3x 4 y 0 3 3 3 y 16 x 0 3x 4 y 0 3 y 16 x 3x 4 y 16 3 y x y x 3 4 Set one equation equal to other to obtain, 16 3 x x 3 4 16 3 x0 3 4 x0 Using equation y 16 x , when x 0 then y 0 , hence the only critical point is 0, 0 . 3 Therefore, the direction of greatest increase (or decrease) in z T x, y when you are at the point 0, 0,80 Hence ant should travel from the coordinate 0, 0,80 Answer 2: a) The contour of f x, y are cos x 3 y k 3 y k cos x 1 1 y k cos x 3 3 Hence the shape of contour of f x, y are, \"graph of cosine function\". Contour of f x, y cos x 3 y b) The gradient of the f x, y is f x, y f x x , y i f y x , y j = sin xi 3 j Therefore direction of greatest increase (steepest uphill) in z , when you are at the point , 2 is given by the direction of 2 f , 2 = sin i 3 j 2 2 = i 3 j , 2 , we will achieve the quickest decent 2 So by pointing in the direction of f (steepest downhill). , 2 i 3j 2 Hence the direction of steepest downhill is f The unit vector in the direction of steepest downhill is i 3 j 2 12 3 1 i 3 j 10 c) , 2 i 3 j , the height would remain 2 If we move perpendicular to the direction of f constant. Hence the direction in which height would remain constant is 3i j The unit vector in which height would remain constant is 3i j 32 12 1 3i j 10 Answer 1: The gradient of the temperature function is T x, y Tx x, y i Ty x, y j = 1 1 3 y 16 x i 3x 4 y j 3 3 To find the critical point, solve Tx x, y 0 and Ty x, y 0 1 3 y 16 x 0 3 3 y 16 x 0 3 y 16 x y 1 3x 4 y 0 3 3x 4 y 0 3x 4 y 16 x 3 y 3 x 4 Set one equation equal to other to obtain, 16 3 x x 3 4 16 3 x 0 3 4 x0 Using equation y 16 x , when x 0 then y 0 , hence the only critical point is 0, 0 . 3 Therefore, the direction of greatest increase (or decrease) in z T x, y when you are at the point 0, 0,80 Hence ant should travel from the coordinate 0, 0,80 Answer 2: a) The contour of f x, y are cos x 3 y k 3 y k cos x 1 1 y k cos x 3 3 Hence the shape of contour of f x, y are, \"graph of cosine function\". Contour of f x, y cos x 3 y b) The gradient of the f x, y is f x, y f x x, y i f y x, y j = sin xi 3j Therefore direction of greatest increase (steepest uphill) in z , when you are at the point , 2 is given by the direction of 2 f , 2 = sin i 3j 2 2 = i 3 j So by pointing in the direction of f , 2 , we will achieve the quickest decent 2 (steepest downhill). Hence the direction of steepest downhill is f , 2 i 3j 2 The unit vector in the direction of steepest downhill is i 3 j 2 12 3 1 i 3 j 10 c) If we move perpendicular to the direction of f , 2 i 3j , the height would remain 2 constant. Hence the direction in which height would remain constant is 3i j The unit vector in which height would remain constant is 3i j 32 12 1 3i j 10 \fAnswer 1: The gradient of the temperature function is T x, y Tx x, y i Ty x, y j 1 1 3 y 16 x i 3x 4 y j 3 3 = To find the critical point, solve Tx x, y 0 and Ty x, y 0 1 1 3 y 16 x 0 3x 4 y 0 3 3 3 y 16 x 0 3x 4 y 0 3 y 16 x 3x 4 y 16 3 y x y x 3 4 Set one equation equal to other to obtain, 16 3 x x 3 4 16 3 x0 3 4 x0 Using equation y 16 x , when x 0 then y 0 , hence the only critical point is 0, 0 . 3 Therefore, the direction of greatest increase (or decrease) in z T x, y when you are at the point 0, 0,80 Hence ant should travel from the coordinate 0, 0,80 Answer 2: a) The contour of f x, y are cos x 3 y k 3 y k cos x 1 1 y k cos x 3 3 Hence the shape of contour of f x, y are, \"graph of cosine function\". Contour of f x, y cos x 3 y b) The gradient of the f x, y is f x, y f x x , y i f y x , y j = sin xi 3 j Therefore direction of greatest increase (steepest uphill) in z , when you are at the point , 2 is given by the direction of 2 f , 2 = sin i 3 j 2 2 = i 3 j , 2 , we will achieve the quickest decent 2 So by pointing in the direction of f (steepest downhill). , 2 i 3j 2 Hence the direction of steepest downhill is f The unit vector in the direction of steepest downhill is i 3 j 2 12 3 1 i 3 j 10 c) , 2 i 3 j , the height would remain 2 If we move perpendicular to the direction of f constant. Hence the direction in which height would remain constant is 3i j The unit vector in which height would remain constant is 3i j 32 12 1 3i j 10 Answer 1: The gradient of the temperature function is T x, y Tx x, y i Ty x, y j = 1 1 3 y 16 x i 3x 4 y j 3 3 To find the critical point, solve Tx x, y 0 and Ty x, y 0 1 3 y 16 x 0 3 3 y 16 x 0 3 y 16 x y 1 3x 4 y 0 3 3x 4 y 0 3x 4 y 16 x 3 y 3 x 4 Set one equation equal to other to obtain, 16 3 x x 3 4 16 3 x 0 3 4 x0 Using equation y 16 x , when x 0 then y 0 , hence the only critical point is 0, 0 . 3 Therefore, the direction of greatest increase (or decrease) in z T x, y when you are at the point 0, 0,80 Hence ant should travel from the coordinate 0, 0,80 Answer 2: a) The contour of f x, y are cos x 3 y k 3 y k cos x 1 1 y k cos x 3 3 Hence the shape of contour of f x, y are, \"graph of cosine function\". Contour of f x, y cos x 3 y b) The gradient of the f x, y is f x, y f x x, y i f y x, y j = sin xi 3j Therefore direction of greatest increase (steepest uphill) in z , when you are at the point , 2 is given by the direction of 2 f , 2 = sin i 3j 2 2 = i 3 j So by pointing in the direction of f , 2 , we will achieve the quickest decent 2 (steepest downhill). Hence the direction of steepest downhill is f , 2 i 3j 2 The unit vector in the direction of steepest downhill is i 3 j 2 12 3 1 i 3 j 10 c) If we move perpendicular to the direction of f , 2 i 3j , the height would remain 2 constant. Hence the direction in which height would remain constant is 3i j The unit vector in which height would remain constant is 3i j 32 12 1 3i j 10