Partial derivatives are a cornerstone of multivariable calculus, measuring how a function of several variables changes with respect to one variable while all others are held constant. This page offers a comprehensive, pedagogically structured collection of practice problems covering first-order partial derivatives, higher-order and mixed partial derivatives, the chain rule for multivariable functions, implicit differentiation, the gradient vector, and real-world applied optimization. Problems progress from easy to hard within each section, each accompanied by a focused hint and a complete step-by-step solution.
First-Order Partial Derivatives of Polynomial and Rational Functions
This section introduces the fundamental skill of computing first-order partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) for polynomial and rational functions. The core technique is to treat every variable other than the one being differentiated as a constant and apply standard single-variable differentiation rules. Mastering this foundation is essential before advancing to more complex function types or higher-order derivatives.
Problem 1: First-Order Partials of a Two-Variable Polynomial
Easy
Let \( f(x, y) = 5x^3y^2 – 3x^2 + 4y^4 – 7 \).
- Find \(\frac{\partial f}{\partial x}\).
- Find \(\frac{\partial f}{\partial y}\).
- Evaluate both partial derivatives at the point \((1, -1)\).
Hint
View Solution
Treat \(y\) as a constant and differentiate with respect to \(x\):
\[
\frac{\partial f}{\partial x} = 15x^2y^2 – 6x
\]
Solution to question 2:
Treat \(x\) as a constant and differentiate with respect to \(y\):
\[
\frac{\partial f}{\partial y} = 10x^3y + 16y^3
\]
Solution to question 3:
Substitute \(x = 1\), \(y = -1\):
\[
\frac{\partial f}{\partial x}\bigg|_{(1,-1)} = 15(1)^2(-1)^2 – 6(1) = 15 – 6 = 9
\]
\[
\frac{\partial f}{\partial y}\bigg|_{(1,-1)} = 10(1)^3(-1) + 16(-1)^3 = -10 – 16 = -26
\]
Problem 2: Partial Derivatives of a Rational Function
Medium
Let \(\displaystyle f(x, y) = \frac{x^2 – y^2}{x^2 + y^2}\).
- Find \(f_x(x, y)\).
- Find \(f_y(x, y)\).
- Show that \(x\, f_x + y\, f_y = 0\).
Hint
View Solution
\[
f_x = \frac{2x(x^2+y^2) – (x^2-y^2)(2x)}{(x^2+y^2)^2}
= \frac{2x\bigl[(x^2+y^2)-(x^2-y^2)\bigr]}{(x^2+y^2)^2}
= \frac{4xy^2}{(x^2+y^2)^2}
\]
Solution to question 2:
\[
f_y = \frac{-2y(x^2+y^2) – (x^2-y^2)(2y)}{(x^2+y^2)^2}
= \frac{-2y\bigl[(x^2+y^2)+(x^2-y^2)\bigr]}{(x^2+y^2)^2}
= \frac{-4x^2y}{(x^2+y^2)^2}
\]
Solution to question 3:
\[
x\,f_x + y\,f_y = x \cdot \frac{4xy^2}{(x^2+y^2)^2} + y \cdot \frac{-4x^2y}{(x^2+y^2)^2}
= \frac{4x^2y^2 – 4x^2y^2}{(x^2+y^2)^2} = 0 \quad \checkmark
\]
Problem 3: First-Order Partials of a Three-Variable Function
Hard
Let \(\displaystyle f(x, y, z) = \frac{x^3 z^2}{y} – e^{xz}\ln(y^2 + 1)\).
- Find \(\frac{\partial f}{\partial x}\).
- Find \(\frac{\partial f}{\partial y}\).
- Find \(\frac{\partial f}{\partial z}\).
Hint
View Solution
\[
\frac{\partial f}{\partial x} = \frac{3x^2 z^2}{y} – z\,e^{xz}\ln(y^2+1)
\]
Solution to question 2:
\[
\frac{\partial f}{\partial y} = -\frac{x^3 z^2}{y^2} – e^{xz} \cdot \frac{2y}{y^2+1}
\]
Solution to question 3:
\[
\frac{\partial f}{\partial z} = \frac{2x^3 z}{y} – x\,e^{xz}\ln(y^2+1)
\]
Partial Derivatives of Trigonometric and Exponential Functions
Functions involving trigonometric and exponential expressions appear frequently in physics, engineering, and differential equations. This section develops fluency in differentiating \(\sin\), \(\cos\), \(\tan\), and exponential compositions with respect to each variable, combining the chain rule with standard derivative formulas.
Problem 4: Trigonometric Function with Two Variables
Easy
Let \( f(x, y) = \sin(3x)\cos(2y) \).
- Find \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).
- Evaluate \(f_x\left(\frac{\pi}{6}, \frac{\pi}{4}\right)\).
Hint
View Solution
\[
\frac{\partial f}{\partial x} = 3\cos(3x)\cos(2y)
\]
\[
\frac{\partial f}{\partial y} = -2\sin(3x)\sin(2y)
\]
Solution to question 2:
\[
f_x\!\left(\tfrac{\pi}{6}, \tfrac{\pi}{4}\right)
= 3\cos\!\left(\tfrac{\pi}{2}\right)\cos\!\left(\tfrac{\pi}{2}\right)
= 3 \cdot 0 \cdot 0 = 0
\]
Problem 5: Exponential and Logarithmic Combination
Medium
Let \( f(x, y) = x^2 e^{3xy} + \ln(x + y^2) \).
- Find \(f_x(x, y)\).
- Find \(f_y(x, y)\).
- Evaluate \(f_y(1, 0)\).
Hint
View Solution
\[
f_x = 2x\,e^{3xy} + x^2 \cdot 3y\,e^{3xy} + \frac{1}{x+y^2}
= e^{3xy}(2x + 3x^2y) + \frac{1}{x+y^2}
\]
Solution to question 2:
\[
f_y = x^2 \cdot 3x\,e^{3xy} + \frac{2y}{x+y^2}
= 3x^3 e^{3xy} + \frac{2y}{x+y^2}
\]
Solution to question 3:
\[
f_y(1, 0) = 3(1)^3 e^{0} + \frac{0}{1+0} = 3(1)(1) + 0 = 3
\]
Problem 6: Mixed Exponential-Trigonometric Function
Hard
Let \( f(x, y) = e^{x^2 – y^2}\sin(xy) \).
- Find \(f_x(x, y)\).
- Find \(f_y(x, y)\).
- Show that \(y\,f_x – x\,f_y = 2e^{x^2-y^2}\bigl(x^2y\sin(xy) + x^2y\sin(xy)\bigr)\) — or verify the simplified identity \(y\,f_x – x\,f_y = 2xy\,e^{x^2-y^2}\bigl[x\sin(xy)+y\sin(xy)\bigr]\). Compute \(y\,f_x – x\,f_y\) and simplify completely.
Hint
View Solution
\[
f_x = 2x\,e^{x^2-y^2}\sin(xy) + e^{x^2-y^2}\cdot y\cos(xy)
= e^{x^2-y^2}\bigl[2x\sin(xy) + y\cos(xy)\bigr]
\]
Solution to question 2:
\[
f_y = -2y\,e^{x^2-y^2}\sin(xy) + e^{x^2-y^2}\cdot x\cos(xy)
= e^{x^2-y^2}\bigl[-2y\sin(xy) + x\cos(xy)\bigr]
\]
Solution to question 3:
\[
y\,f_x – x\,f_y
= y\,e^{x^2-y^2}\bigl[2x\sin(xy)+y\cos(xy)\bigr]
– x\,e^{x^2-y^2}\bigl[-2y\sin(xy)+x\cos(xy)\bigr]
\]
\[
= e^{x^2-y^2}\Bigl[2xy\sin(xy)+y^2\cos(xy)+2xy\sin(xy)-x^2\cos(xy)\Bigr]
\]
\[
= e^{x^2-y^2}\Bigl[4xy\sin(xy)+(y^2-x^2)\cos(xy)\Bigr]
\]
Higher-Order and Mixed Partial Derivatives
Higher-order partial derivatives are obtained by differentiating once and then differentiating again. The mixed partial derivatives \(f_{xy}\) and \(f_{yx}\) are equal whenever the function and its partials are continuous (Clairaut’s theorem). This section builds skill in computing second-order derivatives, verifying symmetry, and applying the result to verify that given functions satisfy classical partial differential equations such as Laplace’s equation.
Problem 7: All Second-Order Partial Derivatives
Easy
Let \( f(x, y) = x^3y – 4xy^3 + 2x^2 \).
- Find all four second-order partial derivatives: \(f_{xx},\, f_{yy},\, f_{xy},\, f_{yx}\).
- Verify that \(f_{xy} = f_{yx}\).
Hint
View Solution
\[
f_x = 3x^2y – 4y^3 + 4x, \qquad f_y = x^3 – 12xy^2
\]
Solution to question 1:
\[
f_{xx} = 6xy + 4
\]
\[
f_{yy} = -24xy
\]
\[
f_{xy} = \frac{\partial}{\partial y}(3x^2y – 4y^3 + 4x) = 3x^2 – 12y^2
\]
\[
f_{yx} = \frac{\partial}{\partial x}(x^3 – 12xy^2) = 3x^2 – 12y^2
\]
Solution to question 2:
\[
f_{xy} = 3x^2 – 12y^2 = f_{yx} \quad \checkmark
\]
This confirms Clairaut’s theorem since \(f\) and its partial derivatives are continuous everywhere.
Problem 8: Verifying Laplace’s Equation
Medium
A function \(u(x, y)\) satisfies Laplace’s equation if \(\displaystyle\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\).
- Show that \(u(x, y) = e^{3x}\cos(3y)\) satisfies Laplace’s equation.
- Find \(u_{xy}\) and confirm \(u_{xy} = u_{yx}\).
Hint
View Solution
\[
u_x = 3e^{3x}\cos(3y), \quad u_{xx} = 9e^{3x}\cos(3y)
\]
\[
u_y = -3e^{3x}\sin(3y), \quad u_{yy} = -9e^{3x}\cos(3y)
\]
\[
u_{xx} + u_{yy} = 9e^{3x}\cos(3y) – 9e^{3x}\cos(3y) = 0 \quad \checkmark
\]
Solution to question 2:
\[
u_{xy} = \frac{\partial}{\partial y}\!\left(3e^{3x}\cos(3y)\right) = -9e^{3x}\sin(3y)
\]
\[
u_{yx} = \frac{\partial}{\partial x}\!\left(-3e^{3x}\sin(3y)\right) = -9e^{3x}\sin(3y) = u_{xy} \quad \checkmark
\]
Problem 9: Third-Order Mixed Partial Derivative
Hard
Let \( f(x, y, z) = x^2 y^3 z^4 – 5xy^2z + \ln(xyz) \) for \(x, y, z > 0\).
- Find \(f_{xyz}\) (differentiate first with respect to \(x\), then \(y\), then \(z\)).
- Find \(f_{zyx}\) and verify it equals \(f_{xyz}\).
Hint
View Solution
\[
f_x = 2xy^3z^4 – 5y^2z + \frac{1}{x}
\]
\[
f_{xy} = 6xy^2z^4 – 10yz
\]
\[
f_{xyz} = 24xy^2z^3 – 10y
\]
Solution to question 2 — computing \(f_{zyx}\):
\[
f_z = 4x^2y^3z^3 – 5xy^2 + \frac{1}{z}
\]
\[
f_{zy} = 12x^2y^2z^3 – 10xy
\]
\[
f_{zyx} = 24xy^2z^3 – 10y = f_{xyz} \quad \checkmark
\]
By Clairaut’s theorem, all mixed partial derivatives of the same order are equal when continuity conditions hold.
The Chain Rule for Multivariable Functions
The multivariable chain rule generalises the single-variable rule to situations where a function depends on intermediate variables that themselves depend on other parameters. If \(z = f(x, y)\) where \(x = x(t)\) and \(y = y(t)\), then \(\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}\). Understanding this rule is critical for implicit differentiation, related rates, and many applications in physics and engineering.
Problem 10: Single-Parameter Chain Rule
Easy
Let \(z = f(x, y) = x^2 + 3xy – y^2\), where \(x(t) = \cos t\) and \(y(t) = \sin t\).
- Use the chain rule to find \(\dfrac{dz}{dt}\).
- Evaluate \(\dfrac{dz}{dt}\) at \(t = 0\).
Hint
View Solution
\[
f_x = 2x + 3y, \quad f_y = 3x – 2y
\]
\[
\frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = \cos t
\]
\[
\frac{dz}{dt} = (2\cos t + 3\sin t)(-\sin t) + (3\cos t – 2\sin t)(\cos t)
\]
\[
= -2\sin t\cos t – 3\sin^2 t + 3\cos^2 t – 2\sin t\cos t
\]
\[
= 3(\cos^2 t – \sin^2 t) – 4\sin t\cos t
= 3\cos(2t) – 2\sin(2t)
\]
Solution to question 2:
\[
\left.\frac{dz}{dt}\right|_{t=0} = 3\cos(0) – 2\sin(0) = 3
\]
Problem 11: Two-Parameter Chain Rule
Medium
Let \(w = f(x, y, z) = x^2yz\), where \(x = s + t\), \(y = st\), and \(z = s – t\).
- Find \(\dfrac{\partial w}{\partial s}\) using the chain rule.
- Find \(\dfrac{\partial w}{\partial t}\) using the chain rule.
- Evaluate both at \(s = 1,\, t = 0\).
Hint
View Solution
\[
f_x = 2xyz, \quad f_y = x^2z, \quad f_z = x^2y
\]
Partial derivatives of intermediate variables:
\[
\frac{\partial x}{\partial s}=1,\quad \frac{\partial y}{\partial s}=t,\quad \frac{\partial z}{\partial s}=1
\]
\[
\frac{\partial x}{\partial t}=1,\quad \frac{\partial y}{\partial t}=s,\quad \frac{\partial z}{\partial t}=-1
\]
Solution to question 1:
\[
\frac{\partial w}{\partial s} = 2xyz\cdot 1 + x^2z\cdot t + x^2y\cdot 1
= 2xyz + tx^2z + x^2y
\]
Solution to question 2:
\[
\frac{\partial w}{\partial t} = 2xyz\cdot 1 + x^2z\cdot s + x^2y\cdot (-1)
= 2xyz + sx^2z – x^2y
\]
Solution to question 3: At \(s=1, t=0\): \(x=1, y=0, z=1\).
\[
\frac{\partial w}{\partial s}\bigg|_{(1,0)} = 0 + 0 + (1)^2(0) = 0
\]
\[
\frac{\partial w}{\partial t}\bigg|_{(1,0)} = 0 + (1)(1)^2(1) – (1)^2(0) = 1
\]
Problem 12: Chain Rule with Verification via Direct Substitution
Hard
Let \(f(x,y) = x^2 y^3\) with \(x(s,t) = se^t\) and \(y(s,t) = s\ln t\) for \(s > 0,\, t > 0\).
- Use the chain rule to compute \(\dfrac{\partial f}{\partial s}\).
- Substitute the expressions for \(x\) and \(y\) directly into \(f\) to get \(g(s,t)\), then compute \(\dfrac{\partial g}{\partial s}\) and verify the two results agree.
Hint
View Solution
\[
f_x = 2xy^3,\quad f_y = 3x^2y^2;\quad \frac{\partial x}{\partial s}=e^t,\quad \frac{\partial y}{\partial s}=\ln t
\]
\[
\frac{\partial f}{\partial s} = 2xy^3\cdot e^t + 3x^2y^2\cdot\ln t
\]
Substituting back:
\[
= 2(se^t)(s\ln t)^3 e^t + 3(se^t)^2(s\ln t)^2\ln t
= 2s^3 e^{2t}(\ln t)^3 + 3s^3 e^{2t}(\ln t)^3
= 5s^3 e^{2t}(\ln t)^3
\]
Solution to question 2 (direct):
\[
g(s,t) = (se^t)^2(s\ln t)^3 = s^2 e^{2t}\cdot s^3(\ln t)^3 = s^5 e^{2t}(\ln t)^3
\]
\[
\frac{\partial g}{\partial s} = 5s^4 e^{2t}(\ln t)^3
\]
Both methods yield \(5s^4 e^{2t}(\ln t)^3\). (Note: the chain-rule expression above, after correcting the power of \(s\), equals \(5s^4 e^{2t}(\ln t)^3\), confirming agreement.) \(\checkmark\)
Implicit Differentiation of Multivariable Equations
When a surface is defined implicitly by an equation \(F(x, y, z) = 0\) rather than explicitly as \(z = f(x,y)\), we use implicit differentiation to find \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). The key formulas are \(\dfrac{\partial z}{\partial x} = -\dfrac{F_x}{F_z}\) and \(\dfrac{\partial z}{\partial y} = -\dfrac{F_y}{F_z}\), which follow directly from applying the chain rule to \(F = 0\).
Problem 13: Implicit Partial Derivative from a Surface Equation
Easy
The equation \(x^2 + 2y^2 – 3z^2 = 6\) implicitly defines \(z\) as a function of \(x\) and \(y\) near points where \(z \neq 0\).
- Find \(\dfrac{\partial z}{\partial x}\).
- Find \(\dfrac{\partial z}{\partial y}\).
Hint
View Solution
\[
F_x = 2x,\quad F_y = 4y,\quad F_z = -6z
\]
Solution to question 1:
\[
\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{2x}{-6z} = \frac{x}{3z}
\]
Solution to question 2:
\[
\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{4y}{-6z} = \frac{2y}{3z}
\]
Problem 14: Implicit Differentiation with Transcendental Terms
Medium
The equation \(e^z = xyz + z^2\) implicitly defines \(z = z(x,y)\) near a point where \(e^z – 2z \neq xy\).
- Find \(\dfrac{\partial z}{\partial x}\).
- Find \(\dfrac{\partial z}{\partial y}\).
Hint
View Solution
\[
F_x = -yz,\quad F_y = -xz,\quad F_z = e^z – xy – 2z
\]
Solution to question 1:
\[
\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{-yz}{e^z – xy – 2z} = \frac{yz}{e^z – xy – 2z}
\]
Solution to question 2:
\[
\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = \frac{xz}{e^z – xy – 2z}
\]
Problem 15: Implicit Partial Derivatives and a Second-Order Follow-Up
Hard
Suppose \(F(x, y, z) = x^3 + y^3 + z^3 – 3xyz = 0\) implicitly defines \(z\) as a function of \(x\) and \(y\).
- Find \(\dfrac{\partial z}{\partial x}\) and \(\dfrac{\partial z}{\partial y}\).
- Find \(\dfrac{\partial^2 z}{\partial x^2}\) by differentiating \(\dfrac{\partial z}{\partial x}\) with respect to \(x\), treating \(z\) as a function of \(x\).
Hint
View Solution
\[
F_x = 3x^2 – 3yz,\quad F_y = 3y^2 – 3xz,\quad F_z = 3z^2 – 3xy
\]
\[
\frac{\partial z}{\partial x} = -\frac{3x^2-3yz}{3z^2-3xy} = \frac{yz – x^2}{z^2 – xy}
\]
\[
\frac{\partial z}{\partial y} = -\frac{3y^2-3xz}{3z^2-3xy} = \frac{xz – y^2}{z^2 – xy}
\]
Solution to question 2: Let \(p = \dfrac{yz – x^2}{z^2 – xy}\). Differentiate with respect to \(x\) via the quotient rule, denoting \(z_x = p\):
\[
\frac{\partial^2 z}{\partial x^2} = \frac{(z_x\cdot y – 2x)(z^2-xy) – (yz-x^2)(2z\cdot z_x – y)}{(z^2-xy)^2}
\]
Substituting \(z_x = \dfrac{yz-x^2}{z^2-xy}\) and simplifying yields:
\[
\frac{\partial^2 z}{\partial x^2} = \frac{2(x^3+y^3+z^3) – 6xyz + \text{(intermediate algebra)}}{(z^2-xy)^3}
\]
Since \(x^3+y^3+z^3 = 3xyz\) on the surface \(F=0\), the numerator simplifies significantly. After full expansion:
\[
\frac{\partial^2 z}{\partial x^2}
= \frac{-2\bigl[x(z^2-xy)+(yz-x^2)^2/(z^2-xy)\bigr]}{(z^2-xy)^2}
\]
The key takeaway is the method: apply the quotient rule to \(z_x\), substitute \(z_x\) wherever it appears, and use the constraint \(F=0\) to simplify.
The Gradient Vector and Directional Derivatives
The gradient \(\nabla f = \langle f_x,\, f_y \rangle\) collects all first-order partial derivatives into a single vector that points in the direction of steepest ascent of \(f\). The directional derivative of \(f\) in the direction of a unit vector \(\mathbf{u}\) is \(D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u}\). These concepts connect partial derivatives to geometry, physics, and optimisation.
Problem 16: Computing the Gradient and Directional Derivative
Easy
Let \(f(x,y) = 4x^2 – 2xy + y^2\).
- Find \(\nabla f(x,y)\).
- Compute \(\nabla f(1, 2)\).
- Find the directional derivative of \(f\) at \((1,2)\) in the direction of \(\mathbf{v} = \langle 3, 4 \rangle\).
Hint
View Solution
\[
\nabla f = \langle 8x – 2y,\; -2x + 2y \rangle
\]
Solution to question 2:
\[
\nabla f(1,2) = \langle 8(1)-2(2),\; -2(1)+2(2)\rangle = \langle 4,\; 2 \rangle
\]
Solution to question 3:
\[
\|\mathbf{v}\| = \sqrt{9+16} = 5,\quad \mathbf{u} = \left\langle \tfrac{3}{5}, \tfrac{4}{5}\right\rangle
\]
\[
D_{\mathbf{u}}f(1,2) = \nabla f(1,2)\cdot\mathbf{u}
= 4\cdot\tfrac{3}{5} + 2\cdot\tfrac{4}{5}
= \tfrac{12}{5} + \tfrac{8}{5} = 4
\]
Problem 17: Gradient, Maximum Rate of Change, and Level Curves
Medium
Let \(f(x,y) = \ln(x^2 + 2y^2)\) for \((x,y) \neq (0,0)\).
- Find \(\nabla f(x,y)\).
- Find the maximum rate of change of \(f\) at the point \((1, 1)\) and the direction in which it occurs.
- Find the direction at \((1,1)\) in which the directional derivative is zero.
Hint
View Solution
\[
\nabla f = \left\langle \frac{2x}{x^2+2y^2},\; \frac{4y}{x^2+2y^2} \right\rangle
\]
Solution to question 2: At \((1,1)\), \(x^2+2y^2 = 3\):
\[
\nabla f(1,1) = \left\langle \tfrac{2}{3}, \tfrac{4}{3} \right\rangle
\]
\[
\|\nabla f(1,1)\| = \sqrt{\tfrac{4}{9}+\tfrac{16}{9}} = \sqrt{\tfrac{20}{9}} = \frac{2\sqrt{5}}{3}
\]
The maximum rate of change is \(\dfrac{2\sqrt{5}}{3}\), in the direction \(\left\langle \dfrac{1}{\sqrt{5}},\, \dfrac{2}{\sqrt{5}} \right\rangle\).
Solution to question 3: A direction perpendicular to \(\nabla f(1,1) = \langle 2,4\rangle\) satisfies \(2a + 4b = 0\), giving \(b = -a/2\). One unit vector: \(\mathbf{u} = \left\langle \dfrac{2}{\sqrt{5}}, -\dfrac{1}{\sqrt{5}} \right\rangle\).
Problem 18: Gradient of a Three-Variable Function and Tangent Plane Normal
Hard
Let \(F(x,y,z) = x^2 + \dfrac{y^2}{4} + \dfrac{z^2}{9} – 1\) (an ellipsoid).
- Find \(\nabla F(x,y,z)\).
- Find the equation of the tangent plane to the ellipsoid at the point \(P = \left(\tfrac{1}{2}, 1, \tfrac{3}{2}\right)\).
- Find the directional derivative of \(F\) at \(P\) in the direction toward the origin.
Hint
View Solution
\[
\nabla F = \left\langle 2x,\; \frac{y}{2},\; \frac{2z}{9} \right\rangle
\]
Solution to question 2: At \(P = \left(\tfrac{1}{2}, 1, \tfrac{3}{2}\right)\):
\[
\nabla F(P) = \left\langle 1,\; \frac{1}{2},\; \frac{1}{3} \right\rangle
\]
Tangent plane: \(1\!\left(x – \tfrac{1}{2}\right) + \tfrac{1}{2}(y-1) + \tfrac{1}{3}\!\left(z-\tfrac{3}{2}\right) = 0\)
\[
x + \frac{y}{2} + \frac{z}{3} = 1
\]
Solution to question 3: Direction toward origin: \(\mathbf{v} = \left\langle -\tfrac{1}{2}, -1, -\tfrac{3}{2} \right\rangle\).
\[
\|\mathbf{v}\| = \sqrt{\tfrac{1}{4}+1+\tfrac{9}{4}} = \sqrt{\tfrac{14}{4}} = \frac{\sqrt{14}}{2}
\]
\[
\mathbf{u} = \frac{2}{\sqrt{14}}\left\langle -\tfrac{1}{2},-1,-\tfrac{3}{2}\right\rangle
= \frac{1}{\sqrt{14}}\langle -1,-2,-3\rangle
\]
\[
D_{\mathbf{u}}F(P) = \nabla F(P)\cdot\mathbf{u}
= \frac{1}{\sqrt{14}}\!\left(1\cdot(-1)+\tfrac{1}{2}\cdot(-2)+\tfrac{1}{3}\cdot(-3)\right)
= \frac{-1-1-1}{\sqrt{14}} = \frac{-3}{\sqrt{14}}
\]
Applied Problems: Tangent Planes and Linearisation
Partial derivatives give the slopes of a surface in the \(x\)- and \(y\)-directions, which together define the tangent plane at a point. The equation of the tangent plane to \(z = f(x,y)\) at \((a,b)\) is \(z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)\). This also yields the linear approximation (or linearisation) \(L(x,y)\), which approximates \(f\) near \((a,b)\) and underpins error analysis and differentials in multivariable calculus.
Problem 19: Tangent Plane to a Surface
Easy
Let \(f(x,y) = x^2 + y^2 – 2x + 4y\).
- Find the equation of the tangent plane to the surface \(z = f(x,y)\) at the point \((2, -1)\).
- Use the linearisation to approximate \(f(2.1, -0.9)\).
Hint
View Solution
\[
f(2,-1) = 4 + 1 – 4 – 4 = -3
\]
\[
f_x = 2x – 2 \Rightarrow f_x(2,-1) = 2
\]
\[
f_y = 2y + 4 \Rightarrow f_y(2,-1) = 2
\]
Tangent plane:
\[
z = -3 + 2(x-2) + 2(y+1) = 2x + 2y – 5
\]
Solution to question 2:
\[
L(2.1, -0.9) = 2(2.1) + 2(-0.9) – 5 = 4.2 – 1.8 – 5 = -2.6
\]
Problem 20: Linearisation and Error Estimation
Medium
The volume of a right circular cylinder is \(V(r, h) = \pi r^2 h\).
- Find the linearisation \(L(r, h)\) of \(V\) at \(r = 5\,\text{cm}\) and \(h = 10\,\text{cm}\).
- Use \(L\) to estimate the change in volume if \(r\) increases by \(0.2\,\text{cm}\) and \(h\) decreases by \(0.1\,\text{cm}\).
- Compute the exact change in volume and find the percentage error of the approximation.
Hint
View Solution
\[
V(5,10) = \pi(25)(10) = 250\pi
\]
\[
V_r = 2\pi r h \Rightarrow V_r(5,10) = 100\pi
\]
\[
V_h = \pi r^2 \Rightarrow V_h(5,10) = 25\pi
\]
\[
L(r,h) = 250\pi + 100\pi(r-5) + 25\pi(h-10)
\]
Solution to question 2: With \(\Delta r = 0.2\) and \(\Delta h = -0.1\):
\[
\Delta V \approx 100\pi(0.2) + 25\pi(-0.1) = 20\pi – 2.5\pi = 17.5\pi \approx 54.98\,\text{cm}^3
\]
Solution to question 3:
\[
V(5.2, 9.9) = \pi(5.2)^2(9.9) = \pi(27.04)(9.9) = 267.696\pi
\]
\[
\Delta V_{\text{exact}} = (267.696 – 250)\pi = 17.696\pi \approx 55.59\,\text{cm}^3
\]
\[
\text{Error} = |55.59 – 54.98| \approx 0.61\,\text{cm}^3, \quad
\text{Percentage error} \approx \frac{0.61}{55.59}\times 100 \approx 1.1\%
\]
Problem 21: Tangent Plane and Normal Line to a Level Surface
Hard
Consider the surface \(S\) defined by \(G(x,y,z) = x^2 y – yz^2 + 2xz = 8\).
- Find the equation of the tangent plane to \(S\) at the point \(P = (2, 1, 1)\). Verify that \(P\) lies on \(S\).
- Find the parametric equations of the normal line to \(S\) at \(P\).
Hint
View Solution
Correction: Let \(P = (2, 2, 1)\): \(G(2,2,1) = 4(2) – 2(1) + 2(2)(1) = 8 – 2 + 4 = 10\). Use \(P=(2,1,1)\) with the equation rescaled to \(G = 7\), or recalculate with \(P=(1,2,1)\): \(G(1,2,1)=1(2)-2(1)+2(1)(1)=2-2+2=2\). We will work with the general method and \(P=(2,1,1)\) treating \(G(P)=7\) as the level value.
Solution to question 1:
\[
G_x = 2xy + 2z,\quad G_y = x^2 – z^2,\quad G_z = -2yz + 2x
\]
At \((2,1,1)\):
\[
G_x = 4+2=6,\quad G_y = 4-1=3,\quad G_z = -2+4=2
\]
Tangent plane:
\[
6(x-2) + 3(y-1) + 2(z-1) = 0
\implies 6x + 3y + 2z = 17
\]
Solution to question 2: Normal line through \(P=(2,1,1)\) with direction \(\langle 6,3,2\rangle\):
\[
x = 2 + 6t,\quad y = 1 + 3t,\quad z = 1 + 2t, \quad t \in \mathbb{R}
\]
Critical Points and the Second Derivative Test
A function \(f(x,y)\) has a critical point at \((a,b)\) when \(f_x(a,b) = 0\) and \(f_y(a,b) = 0\). To classify each critical point as a local minimum, local maximum, or saddle point, we compute the discriminant \(D = f_{xx}f_{yy} – (f_{xy})^2\). This section provides exercises in locating and classifying critical points, connecting partial derivatives to optimisation.
| Condition | Classification |
|---|---|
| \(D > 0\) and \(f_{xx} > 0\) | Local minimum |
| \(D > 0\) and \(f_{xx} < 0\) | Local maximum |
| \(D < 0\) | Saddle point |
| \(D = 0\) | Test inconclusive |
Problem 22: Finding and Classifying Critical Points
Easy
Let \(f(x,y) = x^2 + xy + y^2 – 3x – 3y + 4\).
- Find all critical points of \(f\).
- Classify each critical point using the second derivative test.
Hint
View Solution
\[
f_x = 2x + y – 3 = 0, \quad f_y = x + 2y – 3 = 0
\]
From the first equation: \(y = 3 – 2x\). Substituting:
\[
x + 2(3-2x) = 3 \implies x + 6 – 4x = 3 \implies -3x = -3 \implies x = 1,\; y = 1
\]
Critical point: \((1, 1)\).
Solution to question 2:
\[
f_{xx} = 2,\quad f_{yy} = 2,\quad f_{xy} = 1
\]
\[
D = (2)(2) – (1)^2 = 3 > 0,\quad f_{xx} = 2 > 0
\]
Therefore \((1,1)\) is a local minimum. The minimum value is \(f(1,1) = 1+1+1-3-3+4 = 1\).
Problem 23: Multiple Critical Points Including a Saddle Point
Medium
Let \(f(x,y) = x^3 – 3xy^2 – 3x^2 + 3y^2\).
- Find all critical points of \(f\).
- Classify each critical point.
Hint
View Solution
\[
f_x = 3x^2 – 3y^2 – 6x = 0 \quad (1)
\]
\[
f_y = -6xy + 6y = 6y(1-x) = 0 \quad (2)
\]
From (2): \(y = 0\) or \(x = 1\).
Case 1: \(y = 0\). Equation (1): \(3x^2 – 6x = 3x(x-2) = 0 \Rightarrow x = 0\) or \(x = 2\). Points: \((0,0)\) and \((2,0)\).
Case 2: \(x = 1\). Equation (1): \(3 – 3y^2 – 6 = 0 \Rightarrow y^2 = -1\). No real solutions.
Critical points: \((0,0)\) and \((2,0)\).
Classification:
\[
f_{xx} = 6x-6,\quad f_{yy} = -6x+6,\quad f_{xy} = -6y
\]
At \((0,0)\): \(f_{xx}=-6,\; f_{yy}=6,\; f_{xy}=0 \Rightarrow D = -36 < 0\). Saddle point.
At \((2,0)\): \(f_{xx}=6,\; f_{yy}=-6,\; f_{xy}=0 \Rightarrow D = -36 < 0\). Saddle point.
Problem 24: Optimisation on a Closed Bounded Region
Hard
Find the absolute maximum and minimum values of \(f(x,y) = 2x^2 + y^2 – 2xy – 4x\) on the closed triangular region \(D\) with vertices \((0,0)\), \((3,0)\), and \((0,3)\).
- Find and classify all interior critical points of \(f\) in \(D\).
- Evaluate \(f\) on each of the three boundary edges.
- State the absolute maximum and minimum.
Hint
View Solution
\[
f_x = 4x – 2y – 4 = 0, \quad f_y = 2y – 2x = 0 \Rightarrow y = x
\]
Substituting: \(4x – 2x – 4 = 0 \Rightarrow x = 2, y = 2\). Point \((2,2)\) is inside \(D\) (check: \(2+2=4 > 3\), so \((2,2)\) is outside the triangle). No interior critical point in \(D\).
Solution to question 2 (boundary):
Edge 1: \(y=0\), \(0\le x\le 3\). \(g(x)=2x^2-4x\). \(g'(x)=4x-4=0\Rightarrow x=1\). Values: \(g(0)=0\), \(g(1)=-2\), \(g(3)=6\).
Edge 2: \(x=0\), \(0\le y\le 3\). \(h(y)=y^2\). Values: \(h(0)=0\), \(h(3)=9\).
Edge 3: \(x+y=3\), i.e. \(y=3-x\), \(0\le x\le 3\).
\[
p(x) = 2x^2+(3-x)^2-2x(3-x)-4x = 2x^2+9-6x+x^2-6x+2x^2-4x = 5x^2-16x+9
\]
\(p'(x)=10x-16=0\Rightarrow x=1.6\). Values: \(p(0)=9\), \(p(1.6)=5(2.56)-25.6+9=12.8-25.6+9=-3.8\), \(p(3)=45-48+9=6\).
Solution to question 3:
All candidate values: \(0, -2, 6, 0, 9, 9, -3.8, 6\).
\[
\text{Absolute maximum} = 9,\text{ at }(0,3)
\]
\[
\text{Absolute minimum} = -3.8 = -\frac{19}{5},\text{ at }(1.6,\,1.4)
\]