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Calculus IIIChapter 2: Partial DerivativesSection 2-2

Partial Derivatives — Question 7

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Question 7

For the surface z=f(x,y)=x2+xy+y2,z=f(x,y)=x^2+xy+y^2, hold y=1y=1 fixed and examine the point P=(1,1,3)P=(1,1,3).

Tasks

  1. Find fx(1,1)f_x(1,1).

  2. Obtain the tangent line to the trace y=1y=1 at PP.

  3. Verify the slope by differentiating the trace directly.

Original worksheet page 1: question and worked solution for 2-2-007
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Question 7 – Solution

Strategy. Restrict the surface to the vertical plane y=1y=1; the resulting one-variable curve has slope fx(1,1)f_x(1,1).

See the diagram in the original worksheet below.

Step 1: Partial derivative fx=2x+y,fx(1,1)=3.f_x=2x+y,\qquad \boxed{f_x(1,1)=3}.

Step 2: Trace and tangent At y=1y=1, z=x2+x+1.z=x^2+x+1. The tangent line in the plane y=1y=1 passes through (x,z)=(1,3)(x,z)=(1,3) with slope 33: y=1,z−3=3(x−1).\boxed{y=1,\qquad z-3=3(x-1)}.

Step 3: Direct check Differentiating the trace gives dz/dx=2x+1dz/dx=2x+1, which equals 33 at x=1x=1, matching the partial derivative.

Original worksheet page 2: question and worked solution for 2-2-007

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