(1 point) Consider the function defined by
?(?,?)=??(9?2+5?2)?2+?2F(x,y)=xy(9x2+5y2)x2+y2
except at (?,?)=(0,0)(x,y)=(0,0) where ?(0,0)=0F(0,0)=0.
Then we have
∂∂?∂?∂?(0,0)=∂∂y∂F∂x(0,0)=
∂∂?∂?∂?(0,0)=∂∂x∂F∂y(0,0)=
Note that the answers are different. The existence and continuity
of all second partials in a region around a point guarantees the
equality of the two mixed second derivatives at the point. In the
above case, continuity fails at (0,0)(0,0).
(1 point) Consider the function defined by ?(?,?)=??(9?2+5?2)?2+?2F(x,y)=xy(9x2+5y2)x2+y2 except at (?,?)=(0,0)(x,y)=(0,0) where ?(0,0)=0F(0,0)=0. Then we have...
(1 point) Consider the function defined by F(x, y) = x2 + y2 except at (r, y) - (0, 0) where F(0,0)0 Then we have (0,0) = (0,0) = ax dy Note that the answers are different. The existence and continuity of all second partials in a region around a point guarantees the equality of the two mixed second derivatives at the point. In the above case, continuity fails at (0,0) Note: You can earn partial credit on this problem...
14. (a) Determine all possible critical point(s) of f(x, y) = x2 + xy + y2 – 3x – 6y. (b) without using the Second Order Partial Derivatives Test (SOPDT), de- termine the nature of the obtained C.P(s). (c) Check your answer in (b) through the (SOPDT).
2. Consider the function f : R2 → R defined below. r3уг_ if (x,y) (0,0) f(x,y) = if (x, y) (0, 0) (a) Prove that f is continuous at (0,0) (b) Calculate the partial derivatives (0,0) and (0,0) directly from the definition of partial derivatives. (c) Prove that f is not differentiable at (0,0).
DUE DATE: 23 MARCH 2020 1 1. Let f(x,y) = (x, y) + (0,0) 0. (x, y) = (0,0) evaluate lim(x,y)=(4,3) [5] 2r + 8y 2. Show that lim does not exist. [10] (*.w)-(2,-1) 2.ry + 2 3. Find the first and second partial derivatives of f(x,y) = tan-'(x + 2y). [16] 4. If z is implicitly defined as a function of x and y by I?+y2 + 2 = 1, show az Əz that +y=z [14] ar ду 5....
Observe that the point (1,1,1) satisfies the equation 2. Although we may not be able to write down a formula for z in terms of x and y there is a function z(x,y) that has continuous partial derivatives, is defined for (x,y) near (1,1), and for which z(1,1) 1. For this function find the values of the partials дг/дх (1,1) and дг/ду (1,1). Use this to approximate z(1.1 ,9). Finally, find Эгјах (1,1). If we try to do similar calculations...
Observe that the point (1,1,1) satisfies the equation 2. Although we may not be able to write down a formula for z in terms of x and y there is a function z(x,y) that has continuous partial derivatives, is defined for (x,y) near (1,1), and for which z(1,1)-1. For this function find the values of the partials дг/дх (1,1) and дг/ду (1,1). Use this to approximate z(1.1 ,.9). Finally, find az/0x (1,1). If we try to do similar calculations for...
14. (a) Determine all possible critical point(s) of f(, y) = x2 + xy + y2 - 3.c - 6y. (b) without using the Second Order Partial Derivatives Test (SOPDT), de- termine the nature of the obtained C.P(s). (c) Check your answer in (b) through the (SOPDT). 15. Find a point on the hyperboloid 2z = x2 - y², where the tangent plane is parallel to the plane x - 3y - 2 = 1.
QUESTION 2 Find lim xy +1 (x, y)*(0,0) x2 +y2-1 O-1 O 2 00 O Does not exist.
if (r.y) (0,0), 0,f (, y) (0, 0) 2. Consider f : IR2 -R defined by f(r,y)-+ (a) Show by explicit computation that the directional derivative exists at (x, y)- (0,0) for all oi rections u є R2 with 1 11-1, but that its value %(0.0) (Vf(0,0).u), fr at least one sucli u. (b) Show that the partial derivatives of f are not continuous at (0,0) if (r.y) (0,0), 0,f (, y) (0, 0) 2. Consider f : IR2 -R...
1. Consider XPy4 lim (x,y)=(0,0) x2 + y2 Compute the limit along the two lines y = 0 and y = mx. 2. Let F(x, y) = sin(x2y?), where x = sin(u) + cos(v) and y = eutu. Use the chain rule (substitution will earn zero credit) to find ƏF au