Tangent Planes and Linear Approximations — Question 9

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Question 9

A cylinder has volume V(r,h)=πr2h.V(r,h)=\pi r^2h. Its nominal dimensions are r=10 cmr=10\text{ cm} and h=20 cmh=20\text{ cm}. A measured cylinder instead has Δr=0.03 cm\Delta r=0.03\text{ cm} and Δh=−0.08 cm\Delta h=-0.08\text{ cm}.

Tasks

  1. Use linearization to estimate the change in volume.

  2. Estimate the relative and percentage changes.

  3. Compute the exact change and quantify the linearization error.

Original worksheet page 1: question and worked solution for 3-1-009
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Question 9 – Solution

Strategy. Evaluate the tangent-plane increment ΔV≈VrΔr+VhΔh\Delta V\approx V_r\Delta r+V_h\Delta h at the nominal dimensions.

Step 1: Volume change Vr=2πrh,Vh=πr2.V_r=2\pi rh,\qquad V_h=\pi r^2. At (10,20)(10,20), dV=(400π)(0.03)+(100π)(−0.08)=12π−8π.dV=(400\pi)(0.03)+(100\pi)(-0.08) =12\pi-8\pi. Thus ΔV≈4π cm3≈12.566 cm3.\boxed{\Delta V\approx 4\pi\text{ cm}^3\approx 12.566\text{ cm}^3}.

Step 2: Relative change Dividing the differential by V=πr2hV=\pi r^2h gives dVV=2drr+dhh=20.0310−0.0820=0.002.\frac{dV}{V}=2\frac{dr}{r}+\frac{dh}{h} =2\frac{0.03}{10}-\frac{0.08}{20}=0.002. The estimated percentage increase is .

Step 3: Exact comparison ΔVexact=π(10.03)2(19.92)−2000π=3.969928π cm3.\Delta V_{\mathrm{exact}} =\pi(10.03)^2(19.92)-2000\pi =\boxed{3.969928\pi\text{ cm}^3}. The linear estimate exceeds the exact change by (4−3.969928)π=0.030072π cm3≈0.09447 cm3.(4-3.969928)\pi=\boxed{0.030072\pi\text{ cm}^3\approx 0.09447\text{ cm}^3}.

Original worksheet page 2: question and worked solution for 3-1-009

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