Compute ∇ f \nabla f ∇ f f ( x , y , z ) = ln ( x 2 + y 2 ) + e x z f(x,y,z) = \ln(x^2 + y^2) + e^{xz} f ( x , y , z ) = ln ( x 2 + y 2 ) + e x z ( 1 , 0 , 0 ) (1, 0, 0) ( 1 , 0 , 0 )
Solution f x = 2 x x 2 + y 2 + z e x z f_x = \frac{2x}{x^2+y^2} + ze^{xz} f x = x 2 + y 2 2 x + z e x z f y = 2 y x 2 + y 2 f_y = \frac{2y}{x^2+y^2} f y = x 2 + y 2 2 y f z = x e x z f_z = xe^{xz} f z = x e x z
At ( 1 , 0 , 0 ) (1,0,0) ( 1 , 0 , 0 ) f x = 2 + 0 = 2 f_x = 2 + 0 = 2 f x = 2 + 0 = 2 f y = 0 f_y = 0 f y = 0 f z = 1 f_z = 1 f z = 1
∇ f ( 1 , 0 , 0 ) = ( 2 , 0 , 1 ) \nabla f(1,0,0) = (2, 0, 1) ∇ f ( 1 , 0 , 0 ) = ( 2 , 0 , 1 )
If you get this wrong, revise: Section 1.4 The Gradient.
Let f ( x , y ) = x 3 − 3 x y 2 + y 3 f(x,y) = x^3 - 3xy^2 + y^3 f ( x , y ) = x 3 − 3 x y 2 + y 3
Solution f x = 3 x 2 − 3 y 2 = 0 f_x = 3x^2 - 3y^2 = 0 f x = 3 x 2 − 3 y 2 = 0 f y = − 6 x y + 3 y 2 = 3 y ( − 2 x + y ) = 0 f_y = -6xy + 3y^2 = 3y(-2x + y) = 0 f y = − 6 x y + 3 y 2 = 3 y ( − 2 x + y ) = 0
From f x = 0 f_x = 0 f x = 0 x 2 = y 2 x^2 = y^2 x 2 = y 2 y = ± x y = \pm x y = ± x
If y = x y = x y = x f y = 3 x ( − 2 x + x ) = − 3 x 2 = 0 f_y = 3x(-2x + x) = -3x^2 = 0 f y = 3 x ( − 2 x + x ) = − 3 x 2 = 0 x = 0 x = 0 x = 0 ( 0 , 0 ) (0,0) ( 0 , 0 )
If y = − x y = -x y = − x f y = 3 ( − x ) ( 2 x + x ) = − 9 x 2 = 0 f_y = 3(-x)(2x + x) = -9x^2 = 0 f y = 3 ( − x ) ( 2 x + x ) = − 9 x 2 = 0 x = 0 x = 0 x = 0 ( 0 , 0 ) (0,0) ( 0 , 0 )
The only critical point is ( 0 , 0 ) (0, 0) ( 0 , 0 ) f x x = 6 x f_{xx} = 6x f xx = 6 x f y y = − 6 x + 6 y f_{yy} = -6x + 6y f y y = − 6 x + 6 y f x y = − 6 y f_{xy} = -6y f x y = − 6 y
At ( 0 , 0 ) (0,0) ( 0 , 0 ) D = 0 ⋅ 0 − 0 = 0 D = 0 \cdot 0 - 0 = 0 D = 0 ⋅ 0 − 0 = 0
To classify, note f ( x , y ) = x 3 − 3 x y 2 + y 3 f(x, y) = x^3 - 3xy^2 + y^3 f ( x , y ) = x 3 − 3 x y 2 + y 3 y = 0 y = 0 y = 0 f ( x , 0 ) = x 3 f(x, 0) = x^3 f ( x , 0 ) = x 3 0 0 0 x = y x = y x = y f ( x , x ) = − x 3 f(x, x) = -x^3 f ( x , x ) = − x 3 ( 0 , 0 ) (0, 0) ( 0 , 0 )
If you get this wrong, revise: Section 4.2 Second Derivative Test.
Find the directional derivative of f ( x , y ) = e x cos y f(x,y) = e^x \cos y f ( x , y ) = e x cos y ( 0 , π / 2 ) (0, \pi/2) ( 0 , π /2 ) v = ( 1 , 1 ) \mathbf{v} = (1, 1) v = ( 1 , 1 )
Solution Normalise: ∥ v ∥ = 2 \lVert \mathbf{v} \rVert = \sqrt{2} ∥ v ∥ = 2 u = ( 1 / 2 , 1 / 2 ) \mathbf{u} = (1/\sqrt{2},\, 1/\sqrt{2}) u = ( 1/ 2 , 1/ 2 )
f x = e x cos y f_x = e^x \cos y f x = e x cos y f y = − e x sin y f_y = -e^x \sin y f y = − e x sin y
∇ f ( 0 , π / 2 ) = ( e 0 cos ( π / 2 ) , − e 0 sin ( π / 2 ) ) = ( 0 , − 1 ) \nabla f(0, \pi/2) = (e^0 \cos(\pi/2),\, -e^0 \sin(\pi/2)) = (0, -1) ∇ f ( 0 , π /2 ) = ( e 0 cos ( π /2 ) , − e 0 sin ( π /2 )) = ( 0 , − 1 )
D u f = ( 0 , − 1 ) ⋅ ( 1 / 2 , 1 / 2 ) = − 1 2 D_{\mathbf{u}} f = (0, -1) \cdot (1/\sqrt{2},\, 1/\sqrt{2}) = -\frac{1}{\sqrt{2}} D u f = ( 0 , − 1 ) ⋅ ( 1/ 2 , 1/ 2 ) = − 2 1 If you get this wrong, revise: Section 1.5 Directional Derivatives.
If x 2 z + y 2 z 2 = 5 x^2 z + y^2 z^2 = 5 x 2 z + y 2 z 2 = 5 ∂ z ∂ x \frac{\partial z}{\partial x} ∂ x ∂ z ( 1 , 1 , 1 ) (1, 1, 1) ( 1 , 1 , 1 )
Solution Let F ( x , y , z ) = x 2 z + y 2 z 2 − 5 F(x,y,z) = x^2 z + y^2 z^2 - 5 F ( x , y , z ) = x 2 z + y 2 z 2 − 5 F x = 2 x z F_x = 2xz F x = 2 x z F y = 2 y z 2 F_y = 2yz^2 F y = 2 y z 2 F z = x 2 + 2 y 2 z F_z = x^2 + 2y^2 z F z = x 2 + 2 y 2 z
At ( 1 , 1 , 1 ) (1,1,1) ( 1 , 1 , 1 ) F x = 2 F_x = 2 F x = 2 F z = 1 + 2 = 3 F_z = 1 + 2 = 3 F z = 1 + 2 = 3
∂ z ∂ x = − F x F z = − 2 3 \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{2}{3} ∂ x ∂ z = − F z F x = − 3 2 If you get this wrong, revise: Section 1.9 Implicit Differentiation.
Write the second-order Taylor expansion of f ( x , y ) = sin ( x + y ) f(x,y) = \sin(x + y) f ( x , y ) = sin ( x + y ) ( 0 , 0 ) (0, 0) ( 0 , 0 )
Solution f ( 0 , 0 ) = 0 f(0,0) = 0 f ( 0 , 0 ) = 0 f x = cos ( x + y ) f_x = \cos(x+y) f x = cos ( x + y ) f y = cos ( x + y ) f_y = \cos(x+y) f y = cos ( x + y ) f x ( 0 , 0 ) = f y ( 0 , 0 ) = 1 f_x(0,0) = f_y(0,0) = 1 f x ( 0 , 0 ) = f y ( 0 , 0 ) = 1
f x x = − sin ( x + y ) f_{xx} = -\sin(x+y) f xx = − sin ( x + y ) f x y = − sin ( x + y ) f_{xy} = -\sin(x+y) f x y = − sin ( x + y ) f y y = − sin ( x + y ) f_{yy} = -\sin(x+y) f y y = − sin ( x + y ) f x x ( 0 , 0 ) = f x y ( 0 , 0 ) = f y y ( 0 , 0 ) = 0 f_{xx}(0,0) = f_{xy}(0,0) = f_{yy}(0,0) = 0 f xx ( 0 , 0 ) = f x y ( 0 , 0 ) = f y y ( 0 , 0 ) = 0
f ( x , y ) = 0 + x + y + 1 2 ( 0 ⋅ x 2 + 2 ⋅ 0 ⋅ x y + 0 ⋅ y 2 ) + R 2 = x + y + R 2 f(x,y) = 0 + x + y + \frac{1}{2}(0 \cdot x^2 + 2 \cdot 0 \cdot xy + 0 \cdot y^2) + R_2 = x + y + R_2 f ( x , y ) = 0 + x + y + 2 1 ( 0 ⋅ x 2 + 2 ⋅ 0 ⋅ x y + 0 ⋅ y 2 ) + R 2 = x + y + R 2 Where R 2 = O ( ∣ x ∣ 3 + ∣ y ∣ 3 ) R_2 = O(\lvert x \rvert^3 + \lvert y \rvert^3) R 2 = O (∣ x ∣ 3 + ∣ y ∣ 3 )
If you get this wrong, revise: Section 1.10 Taylor’s Theorem.
Evaluate ∬ D ( x + y ) d A \iint_D (x + y)\, dA ∬ D ( x + y ) d A D D D y = x y = x y = x y = x 2 y = x^2 y = x 2
Solution The curves intersect when x = x 2 x = x^2 x = x 2 x ( x − 1 ) = 0 x(x-1) = 0 x ( x − 1 ) = 0 x = 0 x = 0 x = 0 x = 1 x = 1 x = 1 x ∈ ( 0 , 1 ) x \in (0,1) x ∈ ( 0 , 1 ) x 2 < x x^2 \lt x x 2 < x D = ( x , y ) : 0 ≤ x ≤ 1 , x 2 ≤ y ≤ x D = \\{(x,y) : 0 \leq x \leq 1,\, x^2 \leq y \leq x\\} D = ( x , y ) : 0 ≤ x ≤ 1 , x 2 ≤ y ≤ x
∬ D ( x + y ) d A = ∫ 0 1 ∫ x 2 x ( x + y ) d y d x = ∫ 0 1 [ x y + y 2 2 ] x 2 x d x \iint_D (x + y)\, dA = \int_0^1 \int_{x^2}^x (x + y)\, dy\, dx = \int_0^1 \left[xy + \frac{y^2}{2}\right]_{x^2}^x\, dx ∬ D ( x + y ) d A = ∫ 0 1 ∫ x 2 x ( x + y ) d y d x = ∫ 0 1 [ x y + 2 y 2 ] x 2 x d x = ∫ 0 1 ( x 2 + x 2 2 − x 3 − x 4 2 ) d x = ∫ 0 1 ( 3 x 2 2 − x 3 − x 4 2 ) d x = \int_0^1 \left(x^2 + \frac{x^2}{2} - x^3 - \frac{x^4}{2}\right)\, dx = \int_0^1 \left(\frac{3x^2}{2} - x^3 - \frac{x^4}{2}\right)\, dx = ∫ 0 1 ( x 2 + 2 x 2 − x 3 − 2 x 4 ) d x = ∫ 0 1 ( 2 3 x 2 − x 3 − 2 x 4 ) d x = [ x 3 2 − x 4 4 − x 5 10 ] 0 1 = 1 2 − 1 4 − 1 10 = 10 − 5 − 2 20 = 3 20 = \left[\frac{x^3}{2} - \frac{x^4}{4} - \frac{x^5}{10}\right]_0^1 = \frac{1}{2} - \frac{1}{4} - \frac{1}{10} = \frac{10 - 5 - 2}{20} = \frac{3}{20} = [ 2 x 3 − 4 x 4 − 10 x 5 ] 0 1 = 2 1 − 4 1 − 10 1 = 20 10 − 5 − 2 = 20 3 If you get this wrong, revise: Section 2.2 General Regions.
Evaluate ∭ E x d V \iiint_E x\, dV ∭ E x d V E E E x + y + z = 1 x + y + z = 1 x + y + z = 1
Solution E = ( x , y , z ) : 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 − x , 0 ≤ z ≤ 1 − x − y E = \\{(x,y,z) : 0 \leq x \leq 1,\, 0 \leq y \leq 1-x,\, 0 \leq z \leq 1-x-y\\} E = ( x , y , z ) : 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 − x , 0 ≤ z ≤ 1 − x − y
∭ E x d V = ∫ 0 1 ∫ 0 1 − x ∫ 0 1 − x − y x d z d y d x = ∫ 0 1 ∫ 0 1 − x x ( 1 − x − y ) d y d x \iiint_E x\, dV = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} x\, dz\, dy\, dx = \int_0^1 \int_0^{1-x} x(1-x-y)\, dy\, dx ∭ E x d V = ∫ 0 1 ∫ 0 1 − x ∫ 0 1 − x − y x d z d y d x = ∫ 0 1 ∫ 0 1 − x x ( 1 − x − y ) d y d x = ∫ 0 1 x [ ( 1 − x ) y − y 2 2 ] 0 1 − x d x = ∫ 0 1 x ⋅ ( 1 − x ) 2 2 d x = \int_0^1 x\left[(1-x)y - \frac{y^2}{2}\right]_0^{1-x}\, dx = \int_0^1 x \cdot \frac{(1-x)^2}{2}\, dx = ∫ 0 1 x [ ( 1 − x ) y − 2 y 2 ] 0 1 − x d x = ∫ 0 1 x ⋅ 2 ( 1 − x ) 2 d x = 1 2 ∫ 0 1 x ( 1 − 2 x + x 2 ) d x = 1 2 ∫ 0 1 ( x − 2 x 2 + x 3 ) d x = \frac{1}{2}\int_0^1 x(1 - 2x + x^2)\, dx = \frac{1}{2}\int_0^1 (x - 2x^2 + x^3)\, dx = 2 1 ∫ 0 1 x ( 1 − 2 x + x 2 ) d x = 2 1 ∫ 0 1 ( x − 2 x 2 + x 3 ) d x = 1 2 [ x 2 2 − 2 x 3 3 + x 4 4 ] 0 1 = 1 2 [ 1 2 − 2 3 + 1 4 ] = 1 2 ⋅ 6 − 8 + 3 12 = 1 24 = \frac{1}{2}\left[\frac{x^2}{2} - \frac{2x^3}{3} + \frac{x^4}{4}\right]_0^1 = \frac{1}{2}\left[\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right] = \frac{1}{2} \cdot \frac{6 - 8 + 3}{12} = \frac{1}{24} = 2 1 [ 2 x 2 − 3 2 x 3 + 4 x 4 ] 0 1 = 2 1 [ 2 1 − 3 2 + 4 1 ] = 2 1 ⋅ 12 6 − 8 + 3 = 24 1 If you get this wrong, revise: Section 2.3 Triple Integrals.
Evaluate ∬ D e x 2 + y 2 d A \iint_D e^{x^2+y^2}\, dA ∬ D e x 2 + y 2 d A D = ( x , y ) : 1 ≤ x 2 + y 2 ≤ 4 D = \\{(x,y) : 1 \leq x^2 + y^2 \leq 4\\} D = ( x , y ) : 1 ≤ x 2 + y 2 ≤ 4
Solution Use polar coordinates: 1 ≤ r ≤ 2 1 \leq r \leq 2 1 ≤ r ≤ 2 0 ≤ θ ≤ 2 π 0 \leq \theta \leq 2\pi 0 ≤ θ ≤ 2 π
∬ D e x 2 + y 2 d A = ∫ 0 2 π ∫ 1 2 e r 2 r d r d θ = 2 π ∫ 1 2 r e r 2 d r \iint_D e^{x^2+y^2}\, dA = \int_0^{2\pi} \int_1^2 e^{r^2}\, r\, dr\, d\theta = 2\pi \int_1^2 r e^{r^2}\, dr ∬ D e x 2 + y 2 d A = ∫ 0 2 π ∫ 1 2 e r 2 r d r d θ = 2 π ∫ 1 2 r e r 2 d r Let u = r 2 u = r^2 u = r 2 d u = 2 r d r du = 2r\, dr d u = 2 r d r
= 2 π ⋅ 1 2 ∫ 1 4 e u d u = π ( e 4 − e ) = 2\pi \cdot \frac{1}{2}\int_1^4 e^u\, du = \pi(e^4 - e) = 2 π ⋅ 2 1 ∫ 1 4 e u d u = π ( e 4 − e ) If you get this wrong, revise: Section 2.4 Change of Variables.
Evaluate ∭ E z d V \iiint_E z\, dV ∭ E z d V E E E z ≤ x 2 + y 2 z \leq \sqrt{x^2 + y^2} z ≤ x 2 + y 2 0 ≤ z ≤ 1 0 \leq z \leq 1 0 ≤ z ≤ 1
Solution Use cylindrical coordinates. The cone z = r z = r z = r z = 1 z = 1 z = 1 r = 1 r = 1 r = 1 E ′ = ( r , θ , z ) : 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π , r ≤ z ≤ 1 E' = \\{(r, \theta, z) : 0 \leq r \leq 1,\, 0 \leq \theta \leq 2\pi,\, r \leq z \leq 1\\} E ′ = ( r , θ , z ) : 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π , r ≤ z ≤ 1
∭ E z d V = ∫ 0 2 π ∫ 0 1 ∫ r 1 z r d z d r d θ = 2 π ∫ 0 1 r [ z 2 2 ] r 1 d r \iiint_E z\, dV = \int_0^{2\pi} \int_0^1 \int_r^1 z\, r\, dz\, dr\, d\theta = 2\pi \int_0^1 r\left[\frac{z^2}{2}\right]_r^1\, dr ∭ E z d V = ∫ 0 2 π ∫ 0 1 ∫ r 1 z r d z d r d θ = 2 π ∫ 0 1 r [ 2 z 2 ] r 1 d r = 2 π ∫ 0 1 r 2 ( 1 − r 2 ) d r = π ∫ 0 1 ( r − r 3 ) d r = π [ r 2 2 − r 4 4 ] 0 1 = π ⋅ 1 4 = π 4 = 2\pi \int_0^1 \frac{r}{2}(1 - r^2)\, dr = \pi \int_0^1 (r - r^3)\, dr = \pi\left[\frac{r^2}{2} - \frac{r^4}{4}\right]_0^1 = \pi \cdot \frac{1}{4} = \frac{\pi}{4} = 2 π ∫ 0 1 2 r ( 1 − r 2 ) d r = π ∫ 0 1 ( r − r 3 ) d r = π [ 2 r 2 − 4 r 4 ] 0 1 = π ⋅ 4 1 = 4 π If you get this wrong, revise: Section 2.5 Coordinate System Worked Examples.
Use Green’s theorem to evaluate ∮ C ( 3 y − e sin x ) d x + ( 7 x + y 4 + 1 ) d y \oint_C (3y - e^{\sin x})\, dx + (7x + \sqrt{y^4 + 1})\, dy ∮ C ( 3 y − e s i n x ) d x + ( 7 x + y 4 + 1 ) d y C C C x 2 + y 2 = 9 x^2 + y^2 = 9 x 2 + y 2 = 9
Solution P = 3 y − e sin x P = 3y - e^{\sin x} P = 3 y − e s i n x Q = 7 x + y 4 + 1 Q = 7x + \sqrt{y^4 + 1} Q = 7 x + y 4 + 1
∂ Q ∂ x = 7 , ∂ P ∂ y = 3 \frac{\partial Q}{\partial x} = 7, \quad \frac{\partial P}{\partial y} = 3 ∂ x ∂ Q = 7 , ∂ y ∂ P = 3 By Green’s theorem:
∮ C P d x + Q d y = ∬ D ( 7 − 3 ) d A = 4 ⋅ π ⋅ 9 = 36 π \oint_C P\, dx + Q\, dy = \iint_D (7 - 3)\, dA = 4 \cdot \pi \cdot 9 = 36\pi ∮ C P d x + Q d y = ∬ D ( 7 − 3 ) d A = 4 ⋅ π ⋅ 9 = 36 π If you get this wrong, revise: Section 3.3 Green’s Theorem.
Compute the curl and divergence of F = ( y z , x z , x y ) \mathbf{F} = (yz,\, xz,\, xy) F = ( y z , x z , x y )
Solution Curl:
∇ × F = ( ∂ ( x y ) ∂ y − ∂ ( x z ) ∂ z , ∂ ( y z ) ∂ z − ∂ ( x y ) ∂ x , ∂ ( x z ) ∂ x − ∂ ( y z ) ∂ y ) \nabla \times \mathbf{F} = \left(\frac{\partial (xy)}{\partial y} - \frac{\partial (xz)}{\partial z},\, \frac{\partial (yz)}{\partial z} - \frac{\partial (xy)}{\partial x},\, \frac{\partial (xz)}{\partial x} - \frac{\partial (yz)}{\partial y}\right) ∇ × F = ( ∂ y ∂ ( x y ) − ∂ z ∂ ( x z ) , ∂ z ∂ ( y z ) − ∂ x ∂ ( x y ) , ∂ x ∂ ( x z ) − ∂ y ∂ ( y z ) ) = ( x − x , y − y , z − z ) = 0 = (x - x,\, y - y,\, z - z) = \mathbf{0} = ( x − x , y − y , z − z ) = 0 Divergence:
∇ ⋅ F = ∂ ( y z ) ∂ x + ∂ ( x z ) ∂ y + ∂ ( x y ) ∂ z = 0 + 0 + 0 = 0 \nabla \cdot \mathbf{F} = \frac{\partial (yz)}{\partial x} + \frac{\partial (xz)}{\partial y} + \frac{\partial (xy)}{\partial z} = 0 + 0 + 0 = 0 ∇ ⋅ F = ∂ x ∂ ( y z ) + ∂ y ∂ ( x z ) + ∂ z ∂ ( x y ) = 0 + 0 + 0 = 0 Since the curl is zero and the domain is connected, F \mathbf{F} F F = ∇ ( x y z ) \mathbf{F} = \nabla(xyz) F = ∇ ( x y z )
If you get this wrong, revise: Section 3.4 Curl and Divergence.
Use Stokes’ theorem to evaluate ∮ C F ⋅ d r \oint_C \mathbf{F} \cdot d\mathbf{r} ∮ C F ⋅ d r F = ( 2 y , − z , x ) \mathbf{F} = (2y,\, -z,\, x) F = ( 2 y , − z , x ) C C C x 2 + y 2 = 1 x^2 + y^2 = 1 x 2 + y 2 = 1 z = 1 z = 1 z = 1
Solution Take S S S x 2 + y 2 ≤ 1 x^2 + y^2 \leq 1 x 2 + y 2 ≤ 1 z = 1 z = 1 z = 1 n = ( 0 , 0 , 1 ) \mathbf{n} = (0, 0, 1) n = ( 0 , 0 , 1 )
∇ × F = ( ∂ x ∂ y − ∂ ( − z ) ∂ z , ∂ ( 2 y ) ∂ z − ∂ x ∂ x , ∂ ( − z ) ∂ x − ∂ ( 2 y ) ∂ y ) \nabla \times \mathbf{F} = \left(\frac{\partial x}{\partial y} - \frac{\partial (-z)}{\partial z},\, \frac{\partial (2y)}{\partial z} - \frac{\partial x}{\partial x},\, \frac{\partial (-z)}{\partial x} - \frac{\partial (2y)}{\partial y}\right) ∇ × F = ( ∂ y ∂ x − ∂ z ∂ ( − z ) , ∂ z ∂ ( 2 y ) − ∂ x ∂ x , ∂ x ∂ ( − z ) − ∂ y ∂ ( 2 y ) ) = ( 0 − ( − 1 ) , 0 − 1 , 0 − 2 ) = ( 1 , − 1 , − 2 ) = (0 - (-1),\, 0 - 1,\, 0 - 2) = (1, -1, -2) = ( 0 − ( − 1 ) , 0 − 1 , 0 − 2 ) = ( 1 , − 1 , − 2 ) ( ∇ × F ) ⋅ n = ( 1 , − 1 , − 2 ) ⋅ ( 0 , 0 , 1 ) = − 2 (\nabla \times \mathbf{F}) \cdot \mathbf{n} = (1, -1, -2) \cdot (0, 0, 1) = -2 ( ∇ × F ) ⋅ n = ( 1 , − 1 , − 2 ) ⋅ ( 0 , 0 , 1 ) = − 2 ∮ C F ⋅ d r = ∬ S ( − 2 ) d S = − 2 ⋅ π ⋅ 1 2 = − 2 π \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (-2)\, dS = -2 \cdot \pi \cdot 1^2 = -2\pi ∮ C F ⋅ d r = ∬ S ( − 2 ) d S = − 2 ⋅ π ⋅ 1 2 = − 2 π If you get this wrong, revise: Section 3.5 Stokes’ Theorem.
Use the divergence theorem to compute the flux of F = ( x , y , z ) \mathbf{F} = (x,\, y,\, z) F = ( x , y , z ) [ 0 , 1 ] 3 [0, 1]^3 [ 0 , 1 ] 3
Solution ∇ ⋅ F = ∂ x ∂ x + ∂ y ∂ y + ∂ z ∂ z = 3 \nabla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3 ∇ ⋅ F = ∂ x ∂ x + ∂ y ∂ y + ∂ z ∂ z = 3 ∬ S F ⋅ d S = ∭ E 3 d V = 3 ⋅ 1 3 = 3 \iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E 3\, dV = 3 \cdot 1^3 = 3 ∬ S F ⋅ d S = ∭ E 3 d V = 3 ⋅ 1 3 = 3 If you get this wrong, revise: Section 3.6 Divergence Theorem.
Find a potential function for F = ( 2 x + y , x + 2 z , 2 y ) \mathbf{F} = (2x + y,\, x + 2z,\, 2y) F = ( 2 x + y , x + 2 z , 2 y )
Solution First check: ∇ × F = ( 2 − 2 , 0 − 0 , 1 − 1 ) = 0 \nabla \times \mathbf{F} = (2 - 2,\, 0 - 0,\, 1 - 1) = \mathbf{0} ∇ × F = ( 2 − 2 , 0 − 0 , 1 − 1 ) = 0
∂ ϕ ∂ x = 2 x + y ⟹ ϕ = x 2 + x y + g ( y , z ) \frac{\partial \phi}{\partial x} = 2x + y \implies \phi = x^2 + xy + g(y,z) ∂ x ∂ ϕ = 2 x + y ⟹ ϕ = x 2 + x y + g ( y , z ) ∂ ϕ ∂ y = x + g y = x + 2 z ⟹ g y = 2 z ⟹ g = 2 y z + h ( z ) \frac{\partial \phi}{\partial y} = x + g_y = x + 2z \implies g_y = 2z \implies g = 2yz + h(z) ∂ y ∂ ϕ = x + g y = x + 2 z ⟹ g y = 2 z ⟹ g = 2 y z + h ( z ) ∂ ϕ ∂ z = 2 y + h ′ ( z ) = 2 y ⟹ h ′ ( z ) = 0 ⟹ h ( z ) = C \frac{\partial \phi}{\partial z} = 2y + h'(z) = 2y \implies h'(z) = 0 \implies h(z) = C ∂ z ∂ ϕ = 2 y + h ′ ( z ) = 2 y ⟹ h ′ ( z ) = 0 ⟹ h ( z ) = C ϕ ( x , y , z ) = x 2 + x y + 2 y z + C \phi(x,y,z) = x^2 + xy + 2yz + C ϕ ( x , y , z ) = x 2 + x y + 2 y z + C If you get this wrong, revise: Section 3.7 Conservative Fields and Potential Functions.
Evaluate the surface integral ∬ S ( x 2 + y 2 ) d S \iint_S (x^2 + y^2)\, dS ∬ S ( x 2 + y 2 ) d S S S S x 2 + y 2 = 4 x^2 + y^2 = 4 x 2 + y 2 = 4 0 ≤ z ≤ 3 0 \leq z \leq 3 0 ≤ z ≤ 3
Solution Parametrise the cylinder: r ( θ , z ) = ( 2 cos θ , 2 sin θ , z ) \mathbf{r}(\theta, z) = (2\cos\theta,\, 2\sin\theta,\, z) r ( θ , z ) = ( 2 cos θ , 2 sin θ , z ) 0 ≤ θ ≤ 2 π 0 \leq \theta \leq 2\pi 0 ≤ θ ≤ 2 π 0 ≤ z ≤ 3 0 \leq z \leq 3 0 ≤ z ≤ 3
r θ = ( − 2 sin θ , 2 cos θ , 0 ) \mathbf{r}_\theta = (-2\sin\theta,\, 2\cos\theta,\, 0) r θ = ( − 2 sin θ , 2 cos θ , 0 ) r z = ( 0 , 0 , 1 ) \mathbf{r}_z = (0,\, 0,\, 1) r z = ( 0 , 0 , 1 )
r θ × r z = ( 2 cos θ , 2 sin θ , 0 ) \mathbf{r}_\theta \times \mathbf{r}_z = (2\cos\theta,\, 2\sin\theta,\, 0) r θ × r z = ( 2 cos θ , 2 sin θ , 0 ) ∥ r θ × r z ∥ = 4 cos 2 θ + 4 sin 2 θ = 2 \lVert \mathbf{r}_\theta \times \mathbf{r}_z \rVert = \sqrt{4\cos^2\theta + 4\sin^2\theta} = 2 ∥ r θ × r z ∥ = 4 cos 2 θ + 4 sin 2 θ = 2 On S S S x 2 + y 2 = 4 x^2 + y^2 = 4 x 2 + y 2 = 4
∬ S ( x 2 + y 2 ) d S = ∫ 0 2 π ∫ 0 3 4 ⋅ 2 d z d θ = 8 ⋅ 3 ⋅ 2 π = 48 π \iint_S (x^2 + y^2)\, dS = \int_0^{2\pi} \int_0^3 4 \cdot 2\, dz\, d\theta = 8 \cdot 3 \cdot 2\pi = 48\pi ∬ S ( x 2 + y 2 ) d S = ∫ 0 2 π ∫ 0 3 4 ⋅ 2 d z d θ = 8 ⋅ 3 ⋅ 2 π = 48 π If you get this wrong, revise: Section 5.5 Surface Integrals.
Use Green’s theorem to find the area enclosed by the ellipse x 2 a 2 + y 2 b 2 = 1 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 a 2 x 2 + b 2 y 2 = 1
Solution By Green’s theorem with P = − y / 2 P = -y/2 P = − y /2 Q = x / 2 Q = x/2 Q = x /2
∂ Q ∂ x − ∂ P ∂ y = 1 2 − ( − 1 2 ) = 1 \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1 ∂ x ∂ Q − ∂ y ∂ P = 2 1 − ( − 2 1 ) = 1 So the area is:
A = ∬ D 1 d A = ∮ C − y 2 d x + x 2 d y = 1 2 ∮ C x d y − y d x A = \iint_D 1\, dA = \oint_C -\frac{y}{2}\, dx + \frac{x}{2}\, dy = \frac{1}{2}\oint_C x\, dy - y\, dx A = ∬ D 1 d A = ∮ C − 2 y d x + 2 x d y = 2 1 ∮ C x d y − y d x Parametrise the ellipse: x = a cos t x = a\cos t x = a cos t y = b sin t y = b\sin t y = b sin t 0 ≤ t ≤ 2 π 0 \leq t \leq 2\pi 0 ≤ t ≤ 2 π
A = 1 2 ∫ 0 2 π [ a cos t ⋅ b cos t − b sin t ⋅ ( − a sin t ) ] d t A = \frac{1}{2}\int_0^{2\pi} \left[a\cos t \cdot b\cos t - b\sin t \cdot (-a\sin t)\right]\, dt A = 2 1 ∫ 0 2 π [ a cos t ⋅ b cos t − b sin t ⋅ ( − a sin t ) ] d t = 1 2 ∫ 0 2 π ( a b cos 2 t + a b sin 2 t ) d t = a b 2 ∫ 0 2 π 1 d t = π a b = \frac{1}{2}\int_0^{2\pi} (ab\cos^2 t + ab\sin^2 t)\, dt = \frac{ab}{2}\int_0^{2\pi} 1\, dt = \pi ab = 2 1 ∫ 0 2 π ( ab cos 2 t + ab sin 2 t ) d t = 2 ab ∫ 0 2 π 1 d t = π ab If you get this wrong, revise: Section 3.3 Green’s Theorem.
Find the minimum value of f ( x , y , z ) = x 2 + y 2 + z 2 f(x,y,z) = x^2 + y^2 + z^2 f ( x , y , z ) = x 2 + y 2 + z 2 x + y − z = 1 x + y - z = 1 x + y − z = 1
Solution ∇ f = ( 2 x , 2 y , 2 z ) \nabla f = (2x, 2y, 2z) ∇ f = ( 2 x , 2 y , 2 z ) ∇ g = ( 1 , 1 , − 1 ) \nabla g = (1, 1, -1) ∇ g = ( 1 , 1 , − 1 ) g = x + y − z − 1 g = x + y - z - 1 g = x + y − z − 1
2 x = λ 2x = \lambda 2 x = λ 2 y = λ 2y = \lambda 2 y = λ 2 z = − λ 2z = -\lambda 2 z = − λ x = y = − z x = y = -z x = y = − z
From x + y − z = 1 x + y - z = 1 x + y − z = 1 2 x − ( − x ) = 3 x = 1 2x - (-x) = 3x = 1 2 x − ( − x ) = 3 x = 1 x = 1 / 3 x = 1/3 x = 1/3 y = 1 / 3 y = 1/3 y = 1/3 z = − 1 / 3 z = -1/3 z = − 1/3
f ( 1 / 3 , 1 / 3 , − 1 / 3 ) = 1 9 + 1 9 + 1 9 = 1 3 f(1/3, 1/3, -1/3) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{1}{3} f ( 1/3 , 1/3 , − 1/3 ) = 9 1 + 9 1 + 9 1 = 3 1 This is the minimum (the Hessian of f f f f ≥ 0 f \geq 0 f ≥ 0
If you get this wrong, revise: Section 4.3 Lagrange Multipliers.
Find the arc length of the curve r ( t ) = ( t 2 , 2 t , ln t ) \mathbf{r}(t) = (t^2,\, 2t,\, \ln t) r ( t ) = ( t 2 , 2 t , ln t ) 1 ≤ t ≤ e 1 \leq t \leq e 1 ≤ t ≤ e
Solution r ′ ( t ) = ( 2 t , 2 , 1 / t ) \mathbf{r}'(t) = (2t,\, 2,\, 1/t) r ′ ( t ) = ( 2 t , 2 , 1/ t ) ∥ r ′ ( t ) ∥ = 4 t 2 + 4 + 1 / t 2 \lVert \mathbf{r}'(t) \rVert = \sqrt{4t^2 + 4 + 1/t^2} ∥ r ′ ( t )∥ = 4 t 2 + 4 + 1/ t 2
Note: 4 t 2 + 4 + t − 2 = ( 2 t + 1 / t ) 2 4t^2 + 4 + t^{-2} = (2t + 1/t)^2 4 t 2 + 4 + t − 2 = ( 2 t + 1/ t ) 2 ∥ r ′ ∥ = 2 t + 1 / t \lVert \mathbf{r}' \rVert = 2t + 1/t ∥ r ′ ∥ = 2 t + 1/ t
L = ∫ 1 e ( 2 t + 1 t ) d t = [ t 2 + ln t ] 1 e = e 2 + 1 − 1 − 0 = e 2 L = \int_1^e \left(2t + \frac{1}{t}\right)\, dt = \left[t^2 + \ln t\right]_1^e = e^2 + 1 - 1 - 0 = e^2 L = ∫ 1 e ( 2 t + t 1 ) d t = [ t 2 + ln t ] 1 e = e 2 + 1 − 1 − 0 = e 2 If you get this wrong, revise: Section 5.1 Parametric Curves.
Find the curvature of r ( t ) = ( t , t 2 , t 3 ) \mathbf{r}(t) = (t,\, t^2,\, t^3) r ( t ) = ( t , t 2 , t 3 ) t = 1 t = 1 t = 1
Solution r ′ ( t ) = ( 1 , 2 t , 3 t 2 ) \mathbf{r}'(t) = (1,\, 2t,\, 3t^2) r ′ ( t ) = ( 1 , 2 t , 3 t 2 ) r ′ ′ ( t ) = ( 0 , 2 , 6 t ) \mathbf{r}''(t) = (0,\, 2,\, 6t) r ′′ ( t ) = ( 0 , 2 , 6 t )
At t = 1 t = 1 t = 1 r ′ = ( 1 , 2 , 3 ) \mathbf{r}' = (1, 2, 3) r ′ = ( 1 , 2 , 3 ) r ′ ′ = ( 0 , 2 , 6 ) \mathbf{r}'' = (0, 2, 6) r ′′ = ( 0 , 2 , 6 )
∥ r ′ ∥ = 1 + 4 + 9 = 14 \lVert \mathbf{r}' \rVert = \sqrt{1 + 4 + 9} = \sqrt{14} ∥ r ′ ∥ = 1 + 4 + 9 = 14
r ′ × r ′ ′ = ∣ i j k 1 2 3 0 2 6 ∣ = ( 12 − 6 , − ( 6 − 0 ) , 2 − 0 ) = ( 6 , − 6 , 2 ) \mathbf{r}' \times \mathbf{r}'' = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 0 & 2 & 6 \end{vmatrix} = (12 - 6,\, -(6 - 0),\, 2 - 0) = (6, -6, 2) r ′ × r ′′ = i 1 0 j 2 2 k 3 6 = ( 12 − 6 , − ( 6 − 0 ) , 2 − 0 ) = ( 6 , − 6 , 2 ) ∥ r ′ × r ′ ′ ∥ = 36 + 36 + 4 = 76 = 2 19 \lVert \mathbf{r}' \times \mathbf{r}'' \rVert = \sqrt{36 + 36 + 4} = \sqrt{76} = 2\sqrt{19} ∥ r ′ × r ′′ ∥ = 36 + 36 + 4 = 76 = 2 19 κ = 2 19 ( 14 ) 3 = 2 19 14 14 = 266 98 \kappa = \frac{2\sqrt{19}}{(\sqrt{14})^3} = \frac{2\sqrt{19}}{14\sqrt{14}} = \frac{\sqrt{266}}{98} κ = ( 14 ) 3 2 19 = 14 14 2 19 = 98 266 If you get this wrong, revise: Section 5.2 Curvature and Torsion.
Find the surface area of the part of the sphere x 2 + y 2 + z 2 = 4 x^2 + y^2 + z^2 = 4 x 2 + y 2 + z 2 = 4 z = 1 z = 1 z = 1
Solution Use spherical coordinates. The sphere has ρ = 2 \rho = 2 ρ = 2 z = 1 z = 1 z = 1 2 cos ϕ = 1 2\cos\phi = 1 2 cos ϕ = 1 cos ϕ = 1 / 2 \cos\phi = 1/2 cos ϕ = 1/2 ϕ = π / 3 \phi = \pi/3 ϕ = π /3
The region: 0 ≤ ρ ≤ 2 0 \leq \rho \leq 2 0 ≤ ρ ≤ 2 0 ≤ ϕ ≤ π / 3 0 \leq \phi \leq \pi/3 0 ≤ ϕ ≤ π /3 0 ≤ θ ≤ 2 π 0 \leq \theta \leq 2\pi 0 ≤ θ ≤ 2 π
A = ∫ 0 2 π ∫ 0 π / 3 ρ 2 sin ϕ d ϕ d θ = 4 ⋅ 2 π ∫ 0 π / 3 sin ϕ d ϕ A = \int_0^{2\pi} \int_0^{\pi/3} \rho^2 \sin\phi\, d\phi\, d\theta = 4 \cdot 2\pi \int_0^{\pi/3} \sin\phi\, d\phi A = ∫ 0 2 π ∫ 0 π /3 ρ 2 sin ϕ d ϕ d θ = 4 ⋅ 2 π ∫ 0 π /3 sin ϕ d ϕ = 8 π [ − cos ϕ ] 0 π / 3 = 8 π ( − 1 2 + 1 ) = 8 π ⋅ 1 2 = 4 π = 8\pi \left[-\cos\phi\right]_0^{\pi/3} = 8\pi \left(-\frac{1}{2} + 1\right) = 8\pi \cdot \frac{1}{2} = 4\pi = 8 π [ − cos ϕ ] 0 π /3 = 8 π ( − 2 1 + 1 ) = 8 π ⋅ 2 1 = 4 π If you get this wrong, revise: Section 5.4 Surface Area.
Show that F = ( y e x y + 2 x , x e x y + 2 y ) \mathbf{F} = (ye^{xy} + 2x,\, xe^{xy} + 2y) F = ( y e x y + 2 x , x e x y + 2 y ) ∫ C F ⋅ d r \int_C \mathbf{F} \cdot d\mathbf{r} ∫ C F ⋅ d r C C C ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 1 , 1 ) (1, 1) ( 1 , 1 )
Solution Check: ∂ P ∂ y = e x y + x y e x y \frac{\partial P}{\partial y} = e^{xy} + xye^{xy} ∂ y ∂ P = e x y + x y e x y ∂ Q ∂ x = e x y + x y e x y \frac{\partial Q}{\partial x} = e^{xy} + xye^{xy} ∂ x ∂ Q = e x y + x y e x y F \mathbf{F} F R 2 \mathbb{R}^2 R 2
Find ϕ \phi ϕ
∂ ϕ ∂ x = y e x y + 2 x ⟹ ϕ = e x y + x 2 + g ( y ) \frac{\partial \phi}{\partial x} = ye^{xy} + 2x \implies \phi = e^{xy} + x^2 + g(y) ∂ x ∂ ϕ = y e x y + 2 x ⟹ ϕ = e x y + x 2 + g ( y ) ∂ ϕ ∂ y = x e x y + g ′ ( y ) = x e x y + 2 y ⟹ g ′ ( y ) = 2 y ⟹ g ( y ) = y 2 + C \frac{\partial \phi}{\partial y} = xe^{xy} + g'(y) = xe^{xy} + 2y \implies g'(y) = 2y \implies g(y) = y^2 + C ∂ y ∂ ϕ = x e x y + g ′ ( y ) = x e x y + 2 y ⟹ g ′ ( y ) = 2 y ⟹ g ( y ) = y 2 + C ϕ ( x , y ) = e x y + x 2 + y 2 \phi(x,y) = e^{xy} + x^2 + y^2 ϕ ( x , y ) = e x y + x 2 + y 2 ∫ C F ⋅ d r = ϕ ( 1 , 1 ) − ϕ ( 0 , 0 ) = ( e + 1 + 1 ) − ( 1 + 0 + 0 ) = e + 1 \int_C \mathbf{F} \cdot d\mathbf{r} = \phi(1,1) - \phi(0,0) = (e + 1 + 1) - (1 + 0 + 0) = e + 1 ∫ C F ⋅ d r = ϕ ( 1 , 1 ) − ϕ ( 0 , 0 ) = ( e + 1 + 1 ) − ( 1 + 0 + 0 ) = e + 1 If you get this wrong, revise: Section 3.2 Line Integrals and Section 3.7 Conservative Fields.
Compute the torsion of the curve r ( t ) = ( cosh t , sinh t , t ) \mathbf{r}(t) = (\cosh t,\, \sinh t,\, t) r ( t ) = ( cosh t , sinh t , t ) t = 0 t = 0 t = 0
Solution r ′ ( t ) = ( sinh t , cosh t , 1 ) \mathbf{r}'(t) = (\sinh t,\, \cosh t,\, 1) r ′ ( t ) = ( sinh t , cosh t , 1 ) r ′ ′ ( t ) = ( cosh t , sinh t , 0 ) \mathbf{r}''(t) = (\cosh t,\, \sinh t,\, 0) r ′′ ( t ) = ( cosh t , sinh t , 0 ) r ′ ′ ′ ( t ) = ( sinh t , cosh t , 0 ) \mathbf{r}^{\prime\prime\prime}(t) = (\sinh t,\, \cosh t,\, 0) r ′′′ ( t ) = ( sinh t , cosh t , 0 )
At t = 0 t = 0 t = 0 r ′ = ( 0 , 1 , 1 ) \mathbf{r}' = (0, 1, 1) r ′ = ( 0 , 1 , 1 ) r ′ ′ = ( 1 , 0 , 0 ) \mathbf{r}'' = (1, 0, 0) r ′′ = ( 1 , 0 , 0 ) r ′ ′ ′ = ( 0 , 1 , 0 ) \mathbf{r}^{\prime\prime\prime} = (0, 1, 0) r ′′′ = ( 0 , 1 , 0 )
r ′ × r ′ ′ = ∣ i j k 0 1 1 1 0 0 ∣ = ( 0 , 1 , − 1 ) \mathbf{r}' \times \mathbf{r}'' = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{vmatrix} = (0,\, 1,\, -1) r ′ × r ′′ = i 0 1 j 1 0 k 1 0 = ( 0 , 1 , − 1 ) ∥ r ′ × r ′ ′ ∥ = 2 \lVert \mathbf{r}' \times \mathbf{r}'' \rVert = \sqrt{2} ∥ r ′ × r ′′ ∥ = 2 ( r ′ × r ′ ′ ) ⋅ r ′ ′ ′ = 1 (\mathbf{r}' \times \mathbf{r}'') \cdot \mathbf{r}^{\prime\prime\prime} = 1 ( r ′ × r ′′ ) ⋅ r ′′′ = 1 τ = 1 ( 2 ) 2 = 1 2 \tau = \frac{1}{(\sqrt{2})^2} = \frac{1}{2} τ = ( 2 ) 2 1 = 2 1 If you get this wrong, revise: Section 5.2 Curvature and Torsion.
Evaluate ∭ E 1 x 2 + y 2 + z 2 d V \iiint_E \frac{1}{\sqrt{x^2 + y^2 + z^2}}\, dV ∭ E x 2 + y 2 + z 2 1 d V E E E x 2 + y 2 + z 2 ≤ 1 x^2 + y^2 + z^2 \leq 1 x 2 + y 2 + z 2 ≤ 1
Solution Use spherical coordinates. The integrand is 1 ρ \frac{1}{\rho} ρ 1
∭ E 1 ρ d V = ∫ 0 2 π ∫ 0 π ∫ 0 1 1 ρ ⋅ ρ 2 sin ϕ d ρ d ϕ d θ \iiint_E \frac{1}{\rho}\, dV = \int_0^{2\pi} \int_0^{\pi} \int_0^1 \frac{1}{\rho} \cdot \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta ∭ E ρ 1 d V = ∫ 0 2 π ∫ 0 π ∫ 0 1 ρ 1 ⋅ ρ 2 sin ϕ d ρ d ϕ d θ = ( ∫ 0 1 ρ d ρ ) ( ∫ 0 π sin ϕ d ϕ ) ( ∫ 0 2 π d θ ) = \left(\int_0^1 \rho\, d\rho\right)\left(\int_0^{\pi} \sin\phi\, d\phi\right)\left(\int_0^{2\pi} d\theta\right) = ( ∫ 0 1 ρ d ρ ) ( ∫ 0 π sin ϕ d ϕ ) ( ∫ 0 2 π d θ ) = 1 2 ⋅ 2 ⋅ 2 π = 2 π = \frac{1}{2} \cdot 2 \cdot 2\pi = 2\pi = 2 1 ⋅ 2 ⋅ 2 π = 2 π If you get this wrong, revise: Section 2.5 Coordinate System Worked Examples.
Multivariable calculus extends the ideas of single-variable calculus to higher dimensions. The gradient replaces the derivative as the direction of steepest ascent, and level curves become contour maps of a terrain. Double and triple integrals accumulate quantities over regions and volumes, just as single integrals accumulate over intervals. The major theorems, Green’s, Stokes’, and the divergence theorem, all say the same thing in different costumes: the total circulation or flux through a boundary equals the integral of the curl or divergence inside. This is the fundamental principle that local behaviour determines global outcomes.
Confusing partial and total derivatives. Partial derivatives hold other variables constant; total derivatives account for all variable changes. Fix: ∂ f ∂ x \frac{\partial f}{\partial x} ∂ x ∂ f d f d t = ∂ f ∂ x d x d t + ∂ f ∂ y d y d t \frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} d t df = ∂ x ∂ f d t d x + ∂ y ∂ f d t d y Wrong gradient direction. The gradient ∇ f \nabla f ∇ f Fix: ∇ f \nabla f ∇ f − ∇ f -\nabla f − ∇ f Confusing the Jacobian and Hessian. Jacobian: matrix of first partial derivatives (for transformations). Hessian: matrix of second partial derivatives (for convexity). Fix: Jacobian J i j = ∂ f i ∂ x j J_{ij} = \frac{\partial f_i}{\partial x_j} J ij = ∂ x j ∂ f i H i j = ∂ 2 f ∂ x i ∂ x j H_{ij} = \frac{\partial^2 f}{\partial x_i \partial x_j} H ij = ∂ x i ∂ x j ∂ 2 f Problem. Let f ( x , y ) = x 2 y f(x, y) = x^2 y f ( x , y ) = x 2 y x = cos t x = \cos t x = cos t y = sin t y = \sin t y = sin t d f d t \frac{df}{dt} d t df t = π / 4 t = \pi/4 t = π /4
Solution. d f d t = ∂ f ∂ x d x d t + ∂ f ∂ y d y d t = 2 x y ( − sin t ) + x 2 cos t \frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} = 2xy(-\sin t) + x^2 \cos t d t df = ∂ x ∂ f d t d x + ∂ y ∂ f d t d y = 2 x y ( − sin t ) + x 2 cos t
At t = π / 4 t = \pi/4 t = π /4 x = y = 2 2 x = y = \frac{\sqrt{2}}{2} x = y = 2 2 d f d t = 2 ⋅ 1 2 ⋅ ( − 2 2 ) + 1 2 ⋅ 2 2 = − 2 2 + 2 4 = − 2 4 \frac{df}{dt} = 2 \cdot \frac{1}{2} \cdot (-\frac{\sqrt{2}}{2}) + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{4} = -\frac{\sqrt{2}}{4} d t df = 2 ⋅ 2 1 ⋅ ( − 2 2 ) + 2 1 ⋅ 2 2 = − 2 2 + 4 2 = − 4 2
■ \blacksquare ■
Problem. Evaluate ∬ R x y d A \iint_R x y \, dA ∬ R x y d A R = [ 0 , 1 ] × [ 0 , 2 ] R = [0, 1] \times [0, 2] R = [ 0 , 1 ] × [ 0 , 2 ]
Solution. ∫ 0 1 ∫ 0 2 x y d y d x = ∫ 0 1 x [ y 2 2 ] 0 2 d x = ∫ 0 1 2 x d x = [ x 2 ] 0 1 = 1 \int_0^1 \int_0^2 xy \, dy\, dx = \int_0^1 x \left[\frac{y^2}{2}\right]_0^2 dx = \int_0^1 2x \, dx = [x^2]_0^1 = 1 ∫ 0 1 ∫ 0 2 x y d y d x = ∫ 0 1 x [ 2 y 2 ] 0 2 d x = ∫ 0 1 2 x d x = [ x 2 ] 0 1 = 1
■ \blacksquare ■
Partial derivatives: treat other variables as constants; chain rule for multivariable functions. Gradient ∇ f = ( f x , f y ) \nabla f = (f_x, f_y) ∇ f = ( f x , f y ) ∇ f \nabla f ∇ f Multiple integrals: Fubini’s theorem allows iterated integration; change order with care on bounds. Jacobian determinant: accounts for area/volume scaling under coordinate transformations. Topic Site Link Multivariable Calculus (Overview) WyattsNotes View Real Analysis WyattsNotes View Linear Algebra WyattsNotes View Differential Equations WyattsNotes View Multivariable Calculus, MIT 18.02 MIT OCW View
