Math Problem Statement

the is length of the wire increased by 0.02% and its radius is decrease by 0.005% the percentage change in volume by

Solution

The volume VV of a cylindrical wire is given by:

V=πr2lV = \pi r^2 l

where:

  • rr is the radius of the wire,
  • ll is the length of the wire.

Now, let the percentage change in length be Δl=0.02%\Delta l = 0.02\% and the percentage change in radius be Δr=0.005%\Delta r = -0.005\%.

Step 1: Percentage change in volume formula

The total change in volume can be approximated using the formula for small percentage changes:

ΔVV2Δrr+Δll\frac{\Delta V}{V} \approx 2 \frac{\Delta r}{r} + \frac{\Delta l}{l}

This equation is derived from differentiating the volume formula with respect to rr and ll.

Step 2: Applying the values

  • Δrr=0.005%\frac{\Delta r}{r} = -0.005\%
  • Δll=0.02%\frac{\Delta l}{l} = 0.02\%

Substitute these values into the formula:

ΔVV2(0.005%)+0.02%\frac{\Delta V}{V} \approx 2(-0.005\%) + 0.02\%

ΔVV0.01%+0.02%=0.01%\frac{\Delta V}{V} \approx -0.01\% + 0.02\% = 0.01\%

Final Answer:

The percentage change in volume is 0.01%.

Would you like more details or further clarification?


Here are five related questions for further exploration:

  1. How does the volume change if the radius decreases by a larger percentage?
  2. What happens to the volume if both the radius and length increase?
  3. How does the volume change when only the radius changes?
  4. Can this method be used for other shapes, such as spheres?
  5. What is the significance of small percentage changes in engineering calculations?

Tip: When dealing with small changes, using approximations like linearization makes calculations easier and faster!

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Math Problem Analysis

Mathematical Concepts

Geometry
Calculus
Percentage Change

Formulas

Volume of a cylinder V = πr²l
Percentage change in volume: ΔV/V ≈ 2(Δr/r) + Δl/l

Theorems

Linear Approximation
Volume Formula for Cylinders

Suitable Grade Level

Grades 10-12