Differentials — Question 6

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Question 6

Points constrained to the sphere x2+y2+z2=49x^2+y^2+z^2=49 undergo small changes near (2,3,6)(2,3,6).

Tasks

  1. Derive the differential constraint relating dx,dy,dzdx,dy,dz.

  2. Estimate dzdz when dx=0.03dx=0.03 and dy=−0.02dy=-0.02.

  3. Explain geometrically why the first-order change satisfies this relation.

Original worksheet page 1: question and worked solution for 2-5-006
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Question 6 – Solution

Strategy. Differentiate the constant constraint; its total differential must vanish along allowed infinitesimal changes.

Step 1: Constraint 2xdx+2ydy+2zdz=0,2x\,dx+2y\,dy+2z\,dz=0, so, because z=6≠0z=6\ne 0, dz=−xdx+ydyz.\boxed{dz=-\frac{x\,dx+y\,dy}{z}}.

Step 2: Estimate dz=−2(0.03)+3(−0.02)6=−0.06−0.066=0.dz=-\frac{2(0.03)+3(-0.02)}6 =-\frac{0.06-0.06}{6}=\boxed 0.

Step 3: Geometry The radius vector ⟨x,y,z⟩\langle x,y,z\rangle is normal to the sphere. The relation ⟨x,y,z⟩⋅⟨dx,dy,dz⟩=0\langle x,y,z\rangle\cdot\langle dx,dy,dz\rangle=0 says the first-order displacement is perpendicular to that normal, hence tangent to the sphere. Here the chosen horizontal change is already orthogonal to the radius projection, so no first-order vertical correction is needed.

Original worksheet page 2: question and worked solution for 2-5-006

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