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Calculus IIIChapter 2: Partial DerivativesSection 2-3

Interpretations of Partial Derivatives — Question 2

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Question 2

Temperature is T(x,y)=80−2x2−y2T(x,y)=80-2x^2-y^2 degrees Celsius, with distance in centimeters.

Tasks

  1. Find Tx(3,4),Ty(3,4)T_x(3,4),T_y(3,4) with units.

  2. Interpret signs and magnitudes.

  3. Estimate separate changes for 0.050.05 cm moves in +x+x and +y+y.

Original worksheet page 1: question and worked solution for 2-3-002
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Question 2 – Solution

Strategy. Read each partial as temperature change per centimeter with the other coordinate fixed.

Step 1: Rates Tx=−4xT_x=-4x, Ty=−2yT_y=-2y, so Tx=−12∘C/cm\boxed{T_x=-12\ ^\circ\mathrm C/\mathrm{cm}} and Ty=−8∘C/cm\boxed{T_y=-8\ ^\circ\mathrm C/\mathrm{cm}} at (3,4)(3,4).

Step 2: Meaning Both directions cool; the xx trace is locally steeper.

Step 3: Estimates ΔTx≈−12(.05)=−0.60∘C\Delta T_x\approx-12(.05)=\boxed{-0.60^\circ\mathrm C} and ΔTy≈−8(.05)=−0.40∘C\Delta T_y\approx-8(.05)=\boxed{-0.40^\circ\mathrm C}.

Original worksheet page 2: question and worked solution for 2-3-002

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