Question

There is a possible error of 0.03 cm in a scale of length 1 foot with which the height of a closed right circular cylinder and the diameter of a sphere are measured as 3.5 feet each. If the radii of both cylinder and sphere are same, then the approximate error in the sum of the surface areas of both cylinder and sphere is (in square feet)

• A. 0.385
• B. 0.0962
• C. 0.77
• D. 0.1925

Hint:

For problems on approximate errors, use total differentials. The error $\Delta z$ in a function $z=f(x,y)$ is approximated by $dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$, where $dx$ and $dy$ are the errors in $x$ and $y$. Be mindful of relative vs. absolute errors and unit consistency.

Answer 1

The Correct Option is D

Step 1: Determine the relative error in measurement.
The error is $0.03$ cm for a length of 1 foot. To find the relative error, units must be consistent.
Let's use the approximation $1$ foot $\approx 30$ cm. (This is likely intended for the numbers to work out).
Relative error $E = \frac{\Delta L}{L} = \frac{0.03 \text{ cm}}{30 \text{ cm}} = 0.001$.
This relative error is the same for all measurements made with this scale. So, $\frac{\Delta r}{r} = \frac{\Delta h}{h} = E = 0.001$.
Step 2: Formulate the total surface area.
Sphere radius = Cylinder radius = $r$. Diameter measured is 3.5 ft, so $r = 3.5/2 = 1.75$ ft.
Cylinder height $h$ is measured as 3.5 ft.
Surface area of sphere: $A_s = 4\pi r^2$.
Surface area of closed cylinder: $A_c = 2\pi r^2 + 2\pi rh$.
Total area $S = A_s + A_c = 6\pi r^2 + 2\pi rh$.
Step 3: Find the total differential $dS$ to approximate the error.
$dS = \frac{\partial S}{\partial r} dr + \frac{\partial S}{\partial h} dh$.
$\frac{\partial S}{\partial r} = 12\pi r + 2\pi h$.
$\frac{\partial S}{\partial h} = 2\pi r$.
$dS = (12\pi r + 2\pi h) dr + (2\pi r) dh$.
We know $dr = rE$ and $dh=hE$. Substitute these in:
$dS = (12\pi r + 2\pi h)(rE) + (2\pi r)(hE) = E(12\pi r^2 + 2\pi rh + 2\pi rh) = E(12\pi r^2 + 4\pi rh)$.
Step 4: Substitute the values of $r, h, E$ and $\pi$.
$r = 1.75 = 7/4$ ft. $h = 3.5 = 7/2$ ft. $E = 0.001$. Use $\pi = 22/7$.
$dS = 0.001 \left[ 12\pi (1.75)^2 + 4\pi (1.75)(3.5) \right]$.
$dS = 0.001 \cdot \pi \left[ 12(3.0625) + 4(6.125) \right]$.
$dS = 0.001 \cdot \pi \left[ 36.75 + 24.5 \right] = 0.001 \cdot \pi \cdot (61.25)$.
$dS = 0.06125 \cdot \pi = 0.06125 \cdot \frac{22}{7}$.
$dS = \frac{0.06125 \times 22}{7} = 0.00875 \times 22 = 0.1925$.
The approximate error is $0.1925$ square feet.