The function φ ( x , y ) = x cos ( y 2 ) \varphi(x,y)=x\cos(y^2) is a potential function for F = ⟨ cos ( y 2 ) , – 2 x y sin ( y 2 ) ⟩ \;F=\langle \cos(y^2),\;-2xy\sin(y^2)\rangle Compute ∫ C F · d s \;\int_C F\cdot ds where C is the path made up of straight line segments from P = ( 2 , 0 ) \;P=(2,0) to Q = ( 5 , – 3 π ) Q=(5,-3\pi) to R = ( 3 , π ) R=(3,\sqrt{\pi}) .

Use Lagrange multipliers to determine the maximum value of  f ( x , y ) = x y 2 f(x, y) = xy^2 subject to the constraint x 2 + 2 y 2 = 4 x^2 + 2y^2 = 4 .

Use the chain rule to evaluate ∂ f ∂ r \frac{\partial f}{\partial r} at the point ( r , s ) = ( 3 , – 1 ) (r, s) = (3, -1) where f ( x , y , z ) = x 2 – y z f(x, y, z) = x^2 – yz and x = r + s x = r + s , y = r s y = rs , and z = r 2 + 3 r s z = r^2 + 3rs .

Use the chain rule to evaluate ∂ f ∂ s \frac{\partial f}{\partial s} at the point ( r , s ) = ( 3 , – 1 ) (r, s) = (3, -1) where f ( x , y , z ) = x 2 – y z f(x, y, z) = x^2 – yz and x = r + s , y = r s x = r + s, y = rs , and z = r 2 + 3 r s z = r^2 + 3rs .

Reverse the order of integration to evaluate ∫ 0 4 ∫ y 2 x 3 + 1 d x d y \int_{0}^{4} \int_{\sqrt{y}}^{2} \sqrt{x^3 + 1} \, dx \, dy

Compute the vector line integral ∫ C ⟨ x y z ,   z ,   – x y ⟩ · d s   \int_C \langle xyz,\;z,\;-xy\rangle \cdot ds\; where C is given by r ( t ) = ⟨ 5 t 2 , 1 , t ⟩ , \;r(t)=\langle 5t^2,\;1,\;t\rangle,\; on 0 ≤ t ≤ 1 0\leq t\leq 1 .

Calculate the cross product ⟨ 2 ,   – 1 ,   5 ⟩ × ⟨ 1 ,   – 2 ,   5 ⟩ \langle 2,\ -1,\ 5\rangle \times \langle 1,\ -2,\ 5 \rangle .

Let v = ⟨ 2 , – 1 , 3 ⟩ \mathbf{v} = \langle 2, -1, 3 \rangle and w = ⟨ 1 , 0 , 1 ⟩ \mathbf{w} = \langle 1, 0, 1 \rangle . Calculate the dot product ( 2 v + w ) · ( 3 v – 2 w ) (2\mathbf{v} + \mathbf{w}) \boldsymbol{\cdot} (3\mathbf{v} – 2\mathbf{w}) .

Reverse the order of integration to evaluate ∫ 0 4 ∫ y 2 1 x 3 + 1   d x   d y \int_{0}^{4} \int_{\sqrt{y}}^{2} \frac{1}{\sqrt{x^3 + 1}} \, dx \, dy

Find the directional derivative of f ( x , y ) = x 2 y 2 + 3 x y f(x, y) = x^2y^2 + 3xy in the direction of v = ⟨ 2 , – 1 ⟩ v = \langle 2, -1 \rangle at the point P = ( 3 , 1 ) P = (3, 1) . (Careful: v is not a unit vector.)