Ludium
Sign In
Multivariable Calculus
Vectors and Dot Products
01Vectors, Components, and Length in Space02Scaling and Adding Vectors: The Parallelogram Rule03The Dot Product and the Law of Cosines04Finding Angles in Space With the Dot Product05The Perpendicularity Test and the Normal Vector to a PlaneProblem set0/10Problem set 20/10MIT problem set0/2
Determinants and the Cross Product
01The Component of a Vector Along Any Unit Direction02The 2x2 Determinant and the Area of a Parallelogram03The 3x3 Determinant and the Volume of a Box04The Cross Product and the Right-Hand Rule05The Triple Product and the Volume of a ParallelepipedProblem set0/10Problem set 20/10MIT problem set0/2
Matrices and Inverse Matrices
01Why Swapping a Cross Product Flips Its Sign02The Equation of a Plane Through Three Points03Matrix Multiplication: Rows Dotted With Columns04Matrices as Transformations: Identity and Rotation05The Inverse Matrix and Solving AX = B06Computing a 3x3 Inverse: Minors, Cofactors, AdjointProblem set0/10Problem set 20/10MIT problem set0/2
Planes and Square Systems
01Planes From a Point and a Normal Vector02Reading the Normal Vector Off a Plane Equation03Three Planes, One Point: Solving a 3x3 System04Invertibility, det A, and the Homogeneous System05When det A = 0: Coplanar Normals and a Line of SolutionsProblem set0/10Problem set 20/10
Parametric Lines and Curves
01Parametric Equations of a Line Through Two Points02Where a Line Meets a Plane, and When It Never Does03The Cycloid: Choosing a Parameter That Is Not Time04Deriving the Cycloid's Equations From Three Vectors05Is the Cycloid's Bottom a Corner? Taylor Expansion Finds a CuspProblem set0/10Problem set 20/10MIT problem set0/2
Velocity and Kepler's Law
01Velocity and Speed on the Cycloid: Why the Point Stops Dead02Acceleration Is a Vector: Turning at Constant Speed Still Counts03Arc Length and the Unit Tangent: v = T̂ ds/dt04Equal Areas in Equal Times: Kepler's Second Law and |r × v|05Kepler's Second Law Means the Force Is CentralProblem set0/10Problem set 20/10MIT problem set0/1
Functions of Several Variables
01Functions of Two Variables: Graphs as Surfaces, Built from Slices02Level Curves and Contour Plots: A Surface Drawn as a Flat Map03Reading Contour Plots: Domes, Bowls, Saddles and Mountain Passes04Partial Derivatives: Change One Variable, Hold the Other FixedProblem set0/10Problem set 20/10MIT problem set0/2
Optimization and Least Squares
01Approximation Formula: Changes in x and y Add, via the Tangent Plane02Critical Points: Where Both Partial Derivatives Are Zero03Minimum, Maximum or Saddle Point? Classifying by Completing the Square04Least Squares: The Best-Fit Line as a Minimization in a and b05Solving Least Squares: Best-Fit Lines, Exponentials and ParabolasProblem set0/10Problem set 20/10MIT problem set0/3
The Second Derivative Test
01Quadratic Critical Points: The Sign of 4ac − b² Decides02The Second Derivative Test: Classifying by the Sign of AC − B²03Global Min and Max of x + y + 1/(xy): Boundary and InfinityProblem set0/10Problem set 20/10MIT problem set0/1
Differentials and the Chain Rule
01The Total Differential: Why df Is Not Δf, and Its Three Uses02The Chain Rule: Why It Holds, a Worked Check, and the Product Rule03Chain Rule for a Change of Variables: Partials in Polar CoordinatesProblem set0/10Problem set 20/10MIT problem set0/1
Gradients and Constraints
01The Gradient Vector: Why It Is Perpendicular to Level Sets02Tangent Plane to a Surface From the Gradient: The Hyperboloid03Directional Derivatives and the Direction of Steepest AscentProblem set0/10Problem set 20/10MIT problem set0/2
01Lagrange Multipliers: Tangent Level Curves and ∇f = λ∇g02Solving the Lagrange Equations, and Why the Method Works03The Least-Area Gold Pyramid: Its Top Sits Over the IncenterProblem set0/10Problem set 20/10MIT problem set0/1
01Implicit Partial Derivatives: Differentiate the Constraint02Which Variable Is Held Constant? Subscript Notation03A Constrained Rate Three Ways: Differentials vs Chain RuleProblem set0/10Problem set 20/10MIT problem set0/3
Multiple Integrals
01The Heat Equation: A First Partial Differential Equation02Double Integrals: Volume Under a Surface by Slicing03Iterated Integrals Over a Square and a Quarter Disk04Exchanging the Order of Integration: Redraw the RegionProblem set0/10Problem set 20/10MIT problem set0/1
01Double Integrals in Polar Coordinates: Why dA = r dr dθ02Area, Mass, Averages and Center of Mass by Double Integrals03Moment of Inertia: Spinning a Disk About Its Center and Its RimProblem set0/10Problem set 20/10MIT problem set0/3

Implicit Partial Derivatives: Differentiate the Constraint

How does setting dg = 0 hand you every rate of change among constrained variables, minus sign included, without ever solving for z?


Loading…

←Previous The Least-Area Gold Pyramid: Its Top Sits Over the IncenterNext Which Variable Is Held Constant? Subscript Notation →

Your summary note

    1. 1

      Variables tied by a relation g(x,y,z)=cg(x, y, z) = cg(x,y,z)=c

      Record the setting: variables linked by one relation, such as PV=nRTPV = nRTPV=nRT, with zzz viewed as a function z(x,y)z(x, y)z(x,y) whose partial derivatives are wanted without solving for zzz.

    2. 2

      Setting dg=0dg = 0dg=0 in a worked example

      For x2+yz+z3=8x^2+yz+z^3=8x2+yz+z3=8 at (2,3,1)(2,3,1)(2,3,1), write dgdgdg, set it to zero to get 4 dx+dy+6 dz=04\,dx+dy+6\,dz=04dx+dy+6dz=0, and read off ∂z∂x=−23\frac{\partial z}{\partial x}=-\frac{2}{3}∂x∂z​=−32​ and ∂z∂y=−16\frac{\partial z}{\partial y}=-\frac{1}{6}∂y∂z​=−61​.

    3. 3

      Holding yyy constant as setting dy=0dy = 0dy=0

      Show that putting dy=0dy = 0dy=0 in the example's relation gives the same ∂z∂x\frac{\partial z}{\partial x}∂x∂z​ as the coefficient of dxdxdx, and likewise dx=0dx = 0dx=0 for ∂z∂y\frac{\partial z}{\partial y}∂y∂z​.

    4. 4

      The general formula ∂z∂x=−gxgz\frac{\partial z}{\partial x} = -\frac{g_x}{g_z}∂x∂z​=−gz​gx​​

      Derive it from gx dx+gy dy+gz dz=0g_x\,dx + g_y\,dy + g_z\,dz = 0gx​dx+gy​dy+gz​dz=0 by solving for dzdzdz, write the companion ∂z∂y=−gygz\frac{\partial z}{\partial y} = -\frac{g_y}{g_z}∂y∂z​=−gz​gy​​, and mark the minus sign.

    Attempt 1 of 2