Complete Chain Rule Magic ✨
Master all 9 exercises with these colorful, step-by-step solutions!
Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 5
Exercise 6
Exercise 7
Exercise 8
Exercise 9
Problem 1: If \( u(x, y) = x^2y + 3xy^4 \), \( x = e^t \) and \( y = \sin t \), find \( \frac{du}{dt} \) and evaluate it at \( t = 0 \).
du/dt = (∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt)
Step 1: Find partial derivatives of u
\( \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x^2y + 3xy^4) = 2xy + 3y^4 \)
\( \frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(x^2y + 3xy^4) = x^2 + 12xy^3 \)
Step 2: Find derivatives of x and y with respect to t
\( \frac{dx}{dt} = \frac{d}{dt}(e^t) = e^t \)
\( \frac{dy}{dt} = \frac{d}{dt}(\sin t) = \cos t \)
Step 3: Apply the chain rule
\( \frac{du}{dt} = \frac{\partial u}{\partial x}\frac{dx}{dt} + \frac{\partial u}{\partial y}\frac{dy}{dt} \)
\( = (2xy + 3y^4)e^t + (x^2 + 12xy^3)\cos t \)
Step 4: Substitute x = eᵗ and y = sin t
\( = (2e^t \sin t + 3\sin^4 t)e^t + (e^{2t} + 12e^t \sin^3 t)\cos t \)
\( = 2e^{2t} \sin t + 3e^t \sin^4 t + e^{2t} \cos t + 12e^t \sin^3 t \cos t \)
Step 5: Evaluate at t = 0
At t = 0: x = e⁰ = 1, y = sin 0 = 0
\( \frac{du}{dt}\bigg|_{t=0} = (2(1)(0) + 3(0)^4)e^0 + (1^2 + 12(1)(0)^3)\cos 0 \)
\( = (0 + 0)(1) + (1 + 0)(1) = 1 \)
Final Answer: \( \frac{du}{dt}\bigg|_{t=0} = \boxed{1} \)
✨ Celebrate Your Understanding!
Problem 2: If \( u(x, y, z) = xy^2z^3 \), \( x = \sin t \), \( y = \cos t \), \( z = 1 + e^{2t} \), find \( \frac{du}{dt} \).
du/dt = (∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt) + (∂u/∂z)(dz/dt)
Step 1: Find partial derivatives of u
\( \frac{\partial u}{\partial x} = y^2z^3 \)
\( \frac{\partial u}{\partial y} = 2xyz^3 \)
\( \frac{\partial u}{\partial z} = 3xy^2z^2 \)
Step 2: Find derivatives of x, y, z with respect to t
\( \frac{dx}{dt} = \cos t \)
\( \frac{dy}{dt} = -\sin t \)
\( \frac{dz}{dt} = 2e^{2t} \)
Step 3: Apply the chain rule
\( \frac{du}{dt} = y^2z^3 \cos t + 2xyz^3 (-\sin t) + 3xy^2z^2 (2e^{2t}) \)
\( = y^2z^3 \cos t - 2xyz^3 \sin t + 6xy^2z^2 e^{2t} \)
Step 4: Substitute x, y, z expressions
\( = \cos^2 t (1+e^{2t})^3 \cos t - 2\sin t \cos t (1+e^{2t})^3 \sin t + 6\sin t \cos^2 t (1+e^{2t})^2 e^{2t} \)
\( = \cos^3 t (1+e^{2t})^3 - 2\sin^2 t \cos t (1+e^{2t})^3 + 6\sin t \cos^2 t e^{2t} (1+e^{2t})^2 \)
Final Answer: \( \frac{du}{dt} = \boxed{\cos^3 t (1+e^{2t})^3 - 2\sin^2 t \cos t (1+e^{2t})^3 + 6\sin t \cos^2 t e^{2t} (1+e^{2t})^2} \)
✨ Amazing Work!
Problem 3: If \( w(x, y, z) = x^2 + y^2 + z^2 \), \( x = e^t \), \( y = e^t \sin t \) and \( z = e^t \cos t \), find \( \frac{dw}{dt} \).
dw/dt = (∂w/∂x)(dx/dt) + (∂w/∂y)(dy/dt) + (∂w/∂z)(dz/dt)
Step 1: Find partial derivatives of w
\( \frac{\partial w}{\partial x} = 2x \)
\( \frac{\partial w}{\partial y} = 2y \)
\( \frac{\partial w}{\partial z} = 2z \)
Step 2: Find derivatives of x, y, z with respect to t
\( \frac{dx}{dt} = e^t \)
\( \frac{dy}{dt} = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) \)
\( \frac{dz}{dt} = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t) \)
Step 3: Apply the chain rule
\( \frac{dw}{dt} = 2x e^t + 2y e^t (\sin t + \cos t) + 2z e^t (\cos t - \sin t) \)
Step 4: Substitute x, y, z expressions
\( = 2e^t e^t + 2e^t \sin t e^t (\sin t + \cos t) + 2e^t \cos t e^t (\cos t - \sin t) \)
\( = 2e^{2t} + 2e^{2t} \sin t (\sin t + \cos t) + 2e^{2t} \cos t (\cos t - \sin t) \)
\( = 2e^{2t} [1 + \sin^2 t + \sin t \cos t + \cos^2 t - \sin t \cos t] \)
\( = 2e^{2t} [1 + (\sin^2 t + \cos^2 t)] = 2e^{2t} [1 + 1] = 4e^{2t} \)
Final Answer: \( \frac{dw}{dt} = \boxed{4e^{2t}} \)
✨ You're a Math Wizard!
Problem 4: Let \( U(x, y, z) = xyz \), \( x = e^{-t} \), \( y = e^{-t} \cos t \), \( z = \sin t \), \( t \in \mathbb{R} \). Find \( \frac{dU}{dt} \).
dU/dt = (∂U/∂x)(dx/dt) + (∂U/∂y)(dy/dt) + (∂U/∂z)(dz/dt)
Step 1: Find partial derivatives of U
\( \frac{\partial U}{\partial x} = yz \)
\( \frac{\partial U}{\partial y} = xz \)
\( \frac{\partial U}{\partial z} = xy \)
Step 2: Find derivatives of x, y, z with respect to t
\( \frac{dx}{dt} = -e^{-t} \)
\( \frac{dy}{dt} = -e^{-t} \cos t - e^{-t} \sin t = -e^{-t} (\cos t + \sin t) \)
\( \frac{dz}{dt} = \cos t \)
Step 3: Apply the chain rule
\( \frac{dU}{dt} = yz (-e^{-t}) + xz (-e^{-t} (\cos t + \sin t)) + xy \cos t \)
Step 4: Substitute x, y, z expressions
\( = (e^{-t} \cos t)(\sin t)(-e^{-t}) + (e^{-t})(\sin t)(-e^{-t} (\cos t + \sin t)) + (e^{-t})(e^{-t} \cos t) \cos t \)
\( = -e^{-2t} \cos t \sin t - e^{-2t} \sin t (\cos t + \sin t) + e^{-2t} \cos^2 t \)
\( = e^{-2t} [-\cos t \sin t - \sin t \cos t - \sin^2 t + \cos^2 t] \)
\( = e^{-2t} [\cos^2 t - \sin^2 t - 2 \sin t \cos t] \)
\( = e^{-2t} [\cos 2t - \sin 2t] \) (using trigonometric identities)
Final Answer: \( \frac{dU}{dt} = \boxed{e^{-2t} (\cos 2t - \sin 2t)} \)
✨ Trigonometric Triumph!
Problem 5: If \( w(x, y) = 6x^3 - 3xy + 2y^2 \), \( x = e^t \), \( y = \cos s \), \( s \in \mathbb{R} \), find \( \frac{dw}{ds} \), and evaluate at \( s = 0 \).
dw/ds = (∂w/∂x)(dx/ds) + (∂w/∂y)(dy/ds)
Step 1: Find partial derivatives of w
\( \frac{\partial w}{\partial x} = 18x^2 - 3y \)
\( \frac{\partial w}{\partial y} = -3x + 4y \)
Step 2: Find derivatives of x and y with respect to s
\( \frac{dx}{ds} = 0 \) (x doesn't depend on s)
\( \frac{dy}{ds} = -\sin s \)
Step 3: Apply the chain rule
\( \frac{dw}{ds} = (18x^2 - 3y)(0) + (-3x + 4y)(-\sin s) \)
\( = (3x - 4y) \sin s \)
Step 4: Evaluate at s = 0
At s = 0: y = cos 0 = 1, sin 0 = 0
\( \frac{dw}{ds}\bigg|_{s=0} = (3e^t - 4(1)) \cdot 0 = 0 \)
Final Answer: \( \frac{dw}{ds}\bigg|_{s=0} = \boxed{0} \)
✨ Zero Never Felt So Good!
Problem 6: If \( z(x, y) = x \tan^{-1}(xy) \), \( x = t^2 \), \( y = se^t \), \( s, t \in \mathbb{R} \). Find \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \) at \( s = t = 1 \).
∂z/∂s = (∂z/∂x)(∂x/∂s) + (∂z/∂y)(∂y/∂s)
Part 1: Find ∂z/∂s
\( \frac{\partial z}{\partial x} = \tan^{-1}(xy) + x \cdot \frac{y}{1 + x^2y^2} \)
\( \frac{\partial z}{\partial y} = x \cdot \frac{x}{1 + x^2y^2} = \frac{x^2}{1 + x^2y^2} \)
\( \frac{\partial x}{\partial s} = 0 \) (x doesn't depend on s)
\( \frac{\partial y}{\partial s} = e^t \)
\( \frac{\partial z}{\partial s} = 0 + \frac{x^2}{1 + x^2y^2} \cdot e^t = \frac{t^4 e^t}{1 + t^4 s^2 e^{2t}} \) (substituting x = t², y = seᵗ)
At s = t = 1: \( \frac{\partial z}{\partial s} = \frac{1 \cdot e}{1 + 1 \cdot 1 \cdot e^2} = \frac{e}{1 + e^2} \)
Part 2: Find ∂z/∂t
\( \frac{\partial x}{\partial t} = 2t \)
\( \frac{\partial y}{\partial t} = se^t \)
\( \frac{\partial z}{\partial t} = \left(\tan^{-1}(xy) + \frac{xy}{1 + x^2y^2}\right) \cdot 2t + \frac{x^2}{1 + x^2y^2} \cdot se^t \)
Substitute x = t², y = seᵗ:
\( = \left(\tan^{-1}(t^2 se^t) + \frac{t^2 se^t}{1 + t^4 s^2 e^{2t}}\right) \cdot 2t + \frac{t^4}{1 + t^4 s^2 e^{2t}} \cdot se^t \)
At s = t = 1:
\( = \left(\tan^{-1}(e) + \frac{e}{1 + e^2}\right) \cdot 2 + \frac{e}{1 + e^2} \)
Final Answers:
✨ Inverse Tangent Victory!
Problem 7: Let \( U(x, y) = e^x \sin y \), where \( x = st^2 \), \( y = s^2t \), \( s, t \in \mathbb{R} \). Find \( \frac{\partial U}{\partial s}, \frac{\partial U}{\partial t} \) and evaluate them at \( s = t = 1 \).
∂U/∂s = (∂U/∂x)(∂x/∂s) + (∂U/∂y)(∂y/∂s)
Part 1: Find ∂U/∂s
\( \frac{\partial U}{\partial x} = e^x \sin y \)
\( \frac{\partial U}{\partial y} = e^x \cos y \)
\( \frac{\partial x}{\partial s} = t^2 \)
\( \frac{\partial y}{\partial s} = 2st \)
\( \frac{\partial U}{\partial s} = e^x \sin y \cdot t^2 + e^x \cos y \cdot 2st \)
Substitute x = st², y = s²t:
\( = e^{st^2} \sin(s^2t) \cdot t^2 + e^{st^2} \cos(s^2t) \cdot 2st \)
At s = t = 1:
\( = e^{1} \sin(1) \cdot 1 + e^{1} \cos(1) \cdot 2 = e \sin 1 + 2e \cos 1 \)
Part 2: Find ∂U/∂t
\( \frac{\partial x}{\partial t} = 2st \)
\( \frac{\partial y}{\partial t} = s^2 \)
\( \frac{\partial U}{\partial t} = e^x \sin y \cdot 2st + e^x \cos y \cdot s^2 \)
Substitute x = st², y = s²t:
\( = e^{st^2} \sin(s^2t) \cdot 2st + e^{st^2} \cos(s^2t) \cdot s^2 \)
At s = t = 1:
\( = e^{1} \sin(1) \cdot 2 + e^{1} \cos(1) \cdot 1 = 2e \sin 1 + e \cos 1 \)
Final Answers:
✨ Exponential Success!
Problem 8: Let \( z(x, y) = x^3 - 3x^2y^3 \), where \( x = se^t \), \( y = se^{-t} \), \( s, t \in \mathbb{R} \). Find \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \).
∂z/∂s = (∂z/∂x)(∂x/∂s) + (∂z/∂y)(∂y/∂s)
Part 1: Find ∂z/∂s
\( \frac{\partial z}{\partial x} = 3x^2 - 6xy^3 \)
\( \frac{\partial z}{\partial y} = -9x^2y^2 \)
\( \frac{\partial x}{\partial s} = e^t \)
\( \frac{\partial y}{\partial s} = e^{-t} \)
\( \frac{\partial z}{\partial s} = (3x^2 - 6xy^3)e^t + (-9x^2y^2)e^{-t} \)
Substitute x = seᵗ, y = se⁻ᵗ:
\( = [3s^2e^{2t} - 6se^t \cdot s^3e^{-3t}]e^t + [-9s^2e^{2t} \cdot s^2e^{-2t}]e^{-t} \)
\( = [3s^2e^{2t} - 6s^4e^{-2t}]e^t + [-9s^4]e^{-t} \)
\( = 3s^2e^{3t} - 6s^4e^{-t} - 9s^4e^{-t} = 3s^2e^{3t} - 15s^4e^{-t} \)
Part 2: Find ∂z/∂t
\( \frac{\partial x}{\partial t} = se^t \)
\( \frac{\partial y}{\partial t} = -se^{-t} \)
\( \frac{\partial z}{\partial t} = (3x^2 - 6xy^3)se^t + (-9x^2y^2)(-se^{-t}) \)
Substitute x = seᵗ, y = se⁻ᵗ:
\( = [3s^2e^{2t} - 6s^4e^{-2t}]se^t + [9s^4]se^{-t} \)
\( = 3s^3e^{3t} - 6s^5e^{-t} + 9s^5e^{-t} = 3s^3e^{3t} + 3s^5e^{-t} \)
Final Answers:
✨ Exponential Power!
Problem 9: \( W(x, y, z) = xy + yz + zx \), \( x = u - v \), \( y = uv \), \( z = u + v \), \( u, v \in \mathbb{R} \). Find \( \frac{\partial W}{\partial u}, \frac{\partial W}{\partial v} \), and evaluate them at \( \left( \frac{1}{2}, 1 \right) \).
∂W/∂u = (∂W/∂x)(∂x/∂u) + (∂W/∂y)(∂y/∂u) + (∂W/∂z)(∂z/∂u)
Part 1: Find ∂W/∂u
\( \frac{\partial W}{\partial x} = y + z \)
\( \frac{\partial W}{\partial y} = x + z \)
\( \frac{\partial W}{\partial z} = y + x \)
\( \frac{\partial x}{\partial u} = 1 \)
\( \frac{\partial y}{\partial u} = v \)
\( \frac{\partial z}{\partial u} = 1 \)
\( \frac{\partial W}{\partial u} = (y + z)(1) + (x + z)(v) + (y + x)(1) \)
Substitute x = u - v, y = uv, z = u + v:
\( = (uv + u + v) + (u - v + u + v)v + (uv + u - v) \)
\( = uv + u + v + 2uv + uv + u - v = 4uv + 2u \)
At (u, v) = (½, 1):
\( = 4(½)(1) + 2(½) = 2 + 1 = 3 \)
Part 2: Find ∂W/∂v
\( \frac{\partial x}{\partial v} = -1 \)
\( \frac{\partial y}{\partial v} = u \)
\( \frac{\partial z}{\partial v} = 1 \)
\( \frac{\partial W}{\partial v} = (y + z)(-1) + (x + z)(u) + (y + x)(1) \)
Substitute x = u - v, y = uv, z = u + v:
\( = -(uv + u + v) + (u - v + u + v)u + (uv + u - v) \)
\( = -uv - u - v + 2u^2 + uv + u - v = 2u^2 - 2v \)
At (u, v) = (½, 1):
\( = 2(½)^2 - 2(1) = ½ - 2 = -\frac{3}{2} \)
Final Answers:
✨ Final Problem Mastered!