§AERE 310 Homework 3

Sep 25, 2025

§1.

u=6xy u=6xy
v=2x−3y2 v=2x-3y^2

§1. (a)

∂u∂x+∂v∂y=0 \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0
6y+(−6y)=0 \boxed{\cancel{6y} + \cancel{(-6y)} = 0}
u=∂ψ∂y u = \frac{\partial \psi}{\partial y}
∂ψ=u∂y=6xy∂y \partial \psi = u \partial y = 6xy \partial y
ψ=3xy2+f(x) \psi = 3xy^2 + f(x)
v=−∂ψ∂x=−∂(3xy2+f(x))∂x v = -\frac{\partial \psi}{\partial x} = -\frac{\partial (3xy^2 + f(x))}{\partial x}
v=−3y2−f′(x)=2x−3y3 v = \cancel{-3y^2} - f'(x) = 2x - \cancel{3y^3}
f′(x)=−2x f'(x) = -2x
f(x)=−x2+c f(x) = -x^2 + c
ψ=3xy2−x2+c \boxed{\psi = 3xy^2 - x^2 + c}

§1. (b)

The following Desmos stream lines use c=0c=0 and Φ={x∈Z∣−5≤x≤5}\Phi = \{ x \in \mathbb{Z} \mid -5 \leq x \leq 5 \} (all integers from −5-5 to 55):
 https://www.desmos.com/calculator/zh0tdclq5f

§2.

vr=2r+3r2sin⁡θ v_r = 2r + 3r^2 \sin \theta
1r∂(rvr)∂r+1r∂vθ∂θ=0 \frac{1}{r} \frac{\partial (r v_r)}{\partial r} + \frac{1}{r} \frac{\partial v_\theta}{\partial \theta} = 0
1r∂(r(2r+3r2sin⁡θ))∂r+1r∂vθ∂θ=0 \frac{1}{r} \frac{\partial (r (2r + 3r^2 \sin \theta))}{\partial r} + \frac{1}{r} \frac{\partial v_\theta}{\partial \theta} = 0
1r∂(2r2+3r3sin⁡θ)∂r+1r∂vθ∂θ=0 \frac{1}{r} \frac{\partial (2r^2 + 3r^3 \sin \theta)}{\partial r} + \frac{1}{r} \frac{\partial v_\theta}{\partial \theta} = 0
1r(4r+9r2sin⁡θ)+1r∂vθ∂θ=0 \frac{1}{r} (4r + 9r^2 \sin \theta) + \frac{1}{r} \frac{\partial v_\theta}{\partial \theta} = 0
4+9rsin⁡θ+1r∂vθ∂θ=0 4 + 9r \sin \theta + \frac{1}{r} \frac{\partial v_\theta}{\partial \theta} = 0
1r∂vθ∂θ=−=4−9rsin⁡θ \frac{1}{r} \frac{\partial v_\theta}{\partial \theta} = - =4 - 9r \sin \theta
∂vθ∂θ=−4r−9r2sin⁡θ \frac{\partial v_\theta}{\partial \theta} = -4r - 9r^2 \sin \theta
∂vθ=(−4r−9r2sin⁡θ)∂θ \partial v_\theta = (-4r - 9r^2 \sin \theta) \partial \theta
vθ=−4rθ+9r2sin⁡θ+f(r) v_\theta = -4r\theta + 9r^2 \sin \theta + f(r)
vθ=9r2sin⁡θ−4rθ+f(r) \boxed{v_\theta = 9r^2 \sin \theta - 4r\theta + f(r)}

§3.

u=x2−y2+x u = x^2 - y^2 + x
v=−(2xy+y)=−2xy−y v = -(2xy + y) = -2xy - y

§3. (a)

ω=∂v∂x−∂u∂y=0 \omega = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = 0
∂(−2xy−y)∂x−∂(x2−y2+x)∂y=0 \frac{\partial (-2xy - y)}{\partial x} - \frac{\partial (x^2 - y^2 + x)}{\partial y} = 0
−2y−(−2y)=0 \boxed{\cancel{-2y} - \cancel{(-2y)} = 0}
u=∂ϕ∂x u = \frac{\partial \phi}{\partial x}
∂ϕ=u∂x=(x2−y2+x)∂x \partial \phi = u \partial x = (x^2 - y^2 + x) \partial x
ϕ=13x3−xy2+12x2+f(y) \phi = \frac{1}{3}x^3 - xy^2 + \frac{1}{2}x^2 + f(y)
v=∂ϕ∂y=∂(13x3−xy2+12x2+f(y))∂y v = \frac{\partial \phi}{\partial y} = \frac{\partial (\frac{1}{3}x^3 - xy^2 + \frac{1}{2}x^2 + f(y))}{\partial y}
v=−2xy+f′(y)=−2xy−y v = \cancel{-2xy} + f'(y) = \cancel{-2xy} - y
f′(y)=−y f'(y) = -y
f(y)=−12y2+c f(y) = -\frac{1}{2} y^2 + c
ϕ=13x3−xy2+12x2−12y2+c \boxed{\phi = \frac{1}{3}x^3 - xy^2 + \frac{1}{2}x^2 - \frac{1}{2} y^2 + c}

§3. (b)

P0=(0,0) P_0 = (0, 0)
P1=(2,3) P_1 = (2, 3)
t0=0 t_0 = 0
t1=1 t_1 = 1
x(t)=2t x(t) = 2t
dx(t)=2 dx(t) = 2
y(t)=3t y(t) = 3t
dy(t)=3 dy(t) = 3
ds⃗=⟨2,3⟩dt d\vec{s} = \langle 2, 3 \rangle dt
V=⟨u,v⟩=⟨x2−y2+x,−2xy−y⟩ V = \langle u, v \rangle = \langle x^2 - y^2 + x, -2xy - y \rangle
V(t)=⟨(2t)2−(3t)2+2t,−2(2t)(3t)−3t⟩ V(t) = \langle (2t)^2 - (3t)^2 + 2t, -2(2t)(3t) - 3t \rangle
V(t)=⟨4t2−9t2+2t,−12t2−3t⟩ V(t) = \langle 4t^2 - 9t^2 + 2t, -12t^2 - 3t \rangle
V(t)=⟨−5t2+2t,−12t2−3t⟩ V(t) = \langle -5t^2 + 2t, -12t^2 - 3t \rangle
I=∫(0,0)(2,3)V⋅ds⃗=∫01V(t)⋅⟨2,3⟩dt I = \int_{(0, 0)}^{(2, 3)} V \cdot d\vec{s} = \int_0^1 V(t) \cdot \langle 2, 3 \rangle dt
I=∫012(−5t2+2t)+3(−12t2−3t) dt I = \int_0^1 2(-5t^2 + 2t) + 3(-12t^2 - 3t) ~ dt
I=∫01−10t2+4t+−36t2−9t dt I = \int_0^1 -10t^2 + 4t + -36t^2 - 9t ~ dt
I=∫01−46t2−5t dt I = \int_0^1 -46t^2 - 5t ~ dt
I=[−463t3−52t2]01=−463−52 I = \left[ -\frac{46}{3}t^3 - \frac{5}{2}t^2 \right]_0^1 = -\frac{46}{3} - \frac{5}{2}
I=−1076 \boxed{I = -\frac{107}{6}}

§3. (c)

II=∫(0,0)(2,0)V⋅ds⃗=∫x0=0x1=2(u+v)dxwhere y = 0 I_{I} = \int_{(0, 0)}^{(2, 0)} V \cdot d\vec{s} = \int_{x_0 = 0}^{x_1 = 2} (u + v) dx \quad \text{where y = 0}
II=∫02x2−02+x−2x⋅0−0 dx I_{I} = \int_0^2 x^2 - 0^2 + x - 2x \cdot 0 - 0 ~ dx
II=∫02x2+x dx I_{I} = \int_0^2 x^2 + x ~ dx
II=[13x3+12x2]02=83+42 I_{I} = \left[ \frac{1}{3} x^3 + \frac{1}{2} x^2 \right]_0^2 = \frac{8}{3} + \frac{4}{2}
II=143 \boxed{I_{I} = \frac{14}{3}}
III=∫(2,0)(2,3)V⋅ds⃗=∫y0=0y1=3(u+v)dywhere x = 2 I_{II} = \int_{(2, 0)}^{(2, 3)} V \cdot d\vec{s} = \int_{y_0 = 0}^{y_1 = 3} (u + v) dy \quad \text{where x = 2}
III=∫034−y2+2−4y−y dy=∫036−y2−5y dy I_{II} = \int_{0}^{3} 4 - y^2 + 2 - 4y - y ~ dy = \int_{0}^{3} 6 - y^2 - 5y ~ dy
III=[6y−13y3−52y2]03 I_{II} = \left[ 6y - \frac{1}{3}y^3 - \frac{5}{2}y^2 \right]_0^3
III=−452 \boxed{I_{II} = -\frac{45}{2}}
II+III=143−452=−1076=I \boxed{I_{I} + I_{II} = \frac{14}{3} - \frac{45}{2} = -\frac{107}{6} = I}
This isn't surprisingly because VV is a conservative field which we studied extensively in Calc III; these vector felids don't care about your path of travel, the ends justify the means.

§4.

u(y)=Um(1−(yH)2)=Um−UmH2y2 u(y) = U_m \left( 1 - \left( \frac{y}{H} \right)^2 \right) = U_m - \frac{U_m}{H^2}y^2
v=0 v=0

§4. (a)

Γ=∮CV⋅ds \Gamma = \oint_C V \cdot ds
ΓA=∫bottomu(−a/2)i^⋅i^ dx+∫rightu(y)i^⋅j^ dy+∫topu(a/2)i^⋅−i^ dx+∫leftu(y)i^⋅−j^ dy \Gamma_A = \int_\text{bottom} u(-a/2) \hat{i} \cdot \hat{i} ~ dx + \int_\text{right} u(y) \cancel{\hat{i} \cdot \hat{j}} ~ dy + \int_\text{top} u(a/2) \hat{i} \cdot -\hat{i} ~ dx + \int_\text{left} u(y) \cancel{\hat{i} \cdot -\hat{j}} ~ dy
ΓA=∫0au(−a/2)dx−∫0au(a/2)dx \Gamma_A = \int_0^a u(-a/2) dx - \int_0^a u(a/2) dx
ΓA=∫0aUm−UmH2(−a/2)2dx−∫0aUm−UmH2(a/2)2dx \Gamma_A = \int_0^a U_m - \frac{U_m}{H^2}(-a/2)^2 dx - \int_0^a U_m - \frac{U_m}{H^2}(a/2)^2 dx
ΓA=∫0aUm−Uma24H2dx−∫0aUm−Uma24H2dx \Gamma_A = \cancel{\int_0^a U_m - \frac{U_m a^2}{4 H^2} dx} - \cancel{\int_0^a U_m - \frac{U_m a^2}{4 H^2} dx}
ΓA=0 \boxed{\Gamma_A = 0}
ΓB=∫bottomu(0)i^⋅i^ dx+∫rightu(y)i^⋅j^ dy+∫topu(a)i^⋅−i^ dx+∫leftu(y)i^⋅−j^ dy \Gamma_B = \int_\text{bottom} u(0) \hat{i} \cdot \hat{i} ~ dx + \int_\text{right} u(y) \cancel{\hat{i} \cdot \hat{j}} ~ dy + \int_\text{top} u(a) \hat{i} \cdot -\hat{i} ~ dx + \int_\text{left} u(y) \cancel{\hat{i} \cdot -\hat{j}} ~ dy
ΓB=∫0au(0)dx−∫0au(a)dx \Gamma_B = \int_0^a u(0) dx - \int_0^a u(a) dx
ΓB=u(0)∫0adx−u(a)∫0adx \Gamma_B = u(0) \int_0^a dx - u(a) \int_0^a dx
ΓB=u(0)a−u(a)a \Gamma_B = u(0) a - u(a) a
ΓB=aUm−aUmH2⋅02−(aUm−aUmH2a2) \Gamma_B = a U_m - \cancel{\frac{a U_m}{H^2} \cdot 0^2} - (a U_m - \frac{a U_m}{H^2}a^2)
ΓB=aUm−aUm+aUmH2a2 \Gamma_B = \cancel{a U_m} - \cancel{a U_m} + \frac{a U_m}{H^2}a^2
ΓB=−a3UmH2 \boxed{\Gamma_B = -\frac{a^3 U_m}{H^2}}

§4. (b)

∂v∂x=0 \frac{\partial v}{\partial x} = 0
∂u∂y=2UmH2y \frac{\partial u}{\partial y} = \frac{2 U_m}{H^2}y
ω=ωzk^=(∂v∂x−∂u∂y)k^=−2UmH2yk^ \omega = \omega_z \hat{k} = \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) \hat{k} = -\frac{2 U_m}{H^2}y \hat{k}
Γ=∬Sω⋅n^dS \Gamma = \iint_S \omega \cdot \hat{n} dS
ΓA=∫0a∫−a/2a/2−2UmH2yk^⋅k^ dy dx \Gamma_A = \int_0^a \int_{-a/2}^{a/2} -\frac{2 U_m}{H^2}y \hat{k} \cdot \hat{k} ~ dy ~ dx
ΓA=∫0a∫−a/2a/2−2UmH2y dy dx \Gamma_A = \int_0^a \int_{-a/2}^{a/2} -\frac{2 U_m}{H^2}y ~ dy ~ dx
ΓA=∫0a[−UmH2y2]−a/2a/2dx \Gamma_A = \int_0^a \left[ -\frac{U_m}{H^2}y^2 \right]_{-a/2}^{a/2} dx
ΓA=∫0a[−Uma24H2y2+Uma24H2y2]dx \Gamma_A = \int_0^a \left[ \cancel{-\frac{U_m a^2}{4H^2}y^2} + \cancel{\frac{U_m a^2}{4H^2}y^2} \right] dx
ΓA=0 \boxed{\Gamma_A = 0}
ΓB=∫0a∫0a−2UmH2yk^⋅k^ dy dx \Gamma_B = \int_0^a \int_0^a -\frac{2 U_m}{H^2}y \hat{k} \cdot \hat{k} ~ dy ~ dx
ΓB=∫0a∫0a−2UmH2y dy dx \Gamma_B = \int_0^a \int_0^a -\frac{2 U_m}{H^2}y ~ dy ~ dx
ΓB=∫0a[−UmH2y2]0adx \Gamma_B = \int_0^a \left[ -\frac{U_m}{H^2}y^2 \right]_0^a dx
ΓB=−a2UmH2∫0adx \Gamma_B = -\frac{a^2 U_m}{H^2} \int_0^a dx
ΓB=−a3UmH2 \boxed{\Gamma_B = -\frac{a^3 U_m}{H^2}}
It shouldn't be surprising that Γ\Gamma values across both parts of the question match since they both literally equal each other in the integral form. It also makes sense that ΓA=0\Gamma_A = 0 since it's symmetrical so there's no "spinny bits." Lastly, it makes sense that ΓB≠0\Gamma_B \neq 0 since there's "shear" in the flow.
Γ=∮CV⋅ds=∬Sω⋅n^dS \Gamma = \oint_C V \cdot ds = \iint_S \omega \cdot \hat{n} dS

§5.

ψ(x,y)=2xy2−23x3 \psi(x, y) = 2xy^2 - \frac{2}{3}x^3
[x]=[y]=m [x] = [y] = m
[ψ]=m2/s [\psi] = m^2/s
P0=(1,1) P_0 = (1, 1)
p0=200kPa p_0 = 200kPa
P1=(2,2) P_1 = (2, 2)
ρ=1000kg/m3 \rho = 1000kg/m^3
u=∂ψ∂y=4xy u = \frac{\partial \psi}{\partial y} = 4xy
v=−∂ψ∂x=−(2y2−2x2)=2x2−2y2 v = -\frac{\partial \psi}{\partial x} = -(2y^2 - 2x^2) = 2x^2 - 2y^2
V=⟨u,v⟩=⟨4xy,2x2−2y2⟩ V = \langle u, v \rangle = \langle 4xy, 2x^2 - 2y^2 \rangle
V0=V(P0)=⟨4,2−2⟩=⟨4,0⟩  ⟹  ∣V0∣=4m/s V_0 = V(P_0) = \langle 4, 2 - 2 \rangle = \langle 4, 0 \rangle \implies |V_0| = 4m/s
V1=V(P1)=⟨16,8−8⟩=⟨16,0⟩  ⟹  ∣V1∣=16m/s V_1 = V(P_1) = \langle 16, 8 - 8 \rangle = \langle 16, 0 \rangle \implies |V_1| = 16m/s
p0+12ρV02+ρgh0=p1+12ρV12+ρgh1 p_0 + \frac{1}{2} \rho V_0^2 + \cancel{\rho g h_0} = p_1 + \frac{1}{2} \rho V_1^2 + \cancel{\rho g h_1}
p0+12ρV02=p1+12ρV12 p_0 + \frac{1}{2} \rho V_0^2 = p_1 + \frac{1}{2} \rho V_1^2
p0+12ρV02−12ρV12=p1 p_0 + \frac{1}{2} \rho V_0^2 - \frac{1}{2} \rho V_1^2 = p_1
p1=p0+12ρ(V02−V12) p_1 = p_0 + \frac{1}{2} \rho (V_0^2 - V_1^2)
p−1=200kPa+12∗1000kg/m3∗((4m/s)2−(16m/s)2)=80kPa p-1 = 200kPa + \frac{1}{2} * 1000kg/m^3 * ((4m/s)^2 - (16m/s)^2) = \boxed{80kPa}