(The formula for the surface area of a hemisphere is $A=2 \pi r^{2}$; use 3.14 as an approximation for $\pi$.) Each can of paint costs $\$ 30$ and covers $200 \mathrm{ft}^{2}$.
a) Calculate $\mathrm{dA}$, the approximate difference in the surface area due to the tolerance.
b) Assuming the painters cannot bring partial cans of paint to the job, how many extra cans should they bring to cover the extra area they may encounter?
c) How much extra should the painters plan to spend on paint to account for the possible extra area?

Step 1 ：Given the radius of the hemisphere as 10 ft and the tolerance as 0.1 ft, we first calculate the original surface area of the hemisphere using the formula \(A = 2 \pi r^{2}\), where \(r = 10\) ft. This gives us \(A = 628.32\) ft².

Step 2 ：We then calculate the surface area of the hemisphere with the increased radius (\(r + t\)) using the same formula. This gives us \(A_t = 640.95\) ft².

Step 3 ：The approximate difference in the surface area due to the tolerance (\(dA\)) is then calculated by subtracting the original surface area from the surface area with the increased radius. This gives us \(dA = A_t - A = 12.63\) ft².

Step 4 ：We then calculate the number of extra cans of paint needed by dividing the extra surface area (\(dA\)) by the area that one can of paint covers (200 ft²). Since we cannot bring partial cans of paint, we round this number up to the nearest whole number. This gives us 1 extra can of paint.

Step 5 ：Finally, we calculate the extra cost by multiplying the number of extra cans of paint by the cost of one can of paint ($30). This gives us an extra cost of $30.

Step 6 ：Final Answer: a) The approximate difference in the surface area due to the tolerance is \(\boxed{12.63 \, \text{ft}^2}\). b) The painters should bring \(\boxed{1}\) extra can of paint to cover the extra area they may encounter. c) The painters should plan to spend an extra \(\boxed{\$30}\) on paint to account for the possible extra area.