Question

In a previous exercise we discussed a water wheel of radius 15ft which is submerged 3ft in flowing water. A beetle is sitting on the very edge of the water wheel and we found that when the beetle made an angle of θ with the horizontal the beetle was f(θ)ft above the water where f(θ)=15sin(θ)+12. Assuming that θ is in radians now (the same equation still works), find the angle θ so that the beetle is exactly at the surface of the water, so that 8π≤θ≤10π, and so that cos(θ)>0. Round to two decimal places.

Answer 1

To find the angle θ so that the beetle is exactly at the surface of the water, set the function f(θ) equal to 0 and solve for θ. The angle θ is approximately 9.42 radians.

Step-by-step explanation:

To find the angle θ so that the beetle is exactly at the surface of the water, we need to set the function f(θ) equal to 0, as the beetle would be at water level. The equation is given by f(θ) = 15sin(θ) + 12. Setting f(θ) = 0, we have 15sin(θ) + 12 = 0. Rearranging the equation, we find that sin(θ) = -12/15 = -0.8.

Since we want the angle θ such that 8π≤θ≤10π, we need to find the value of θ in that range which satisfies sin(θ) = -0.8. Using a calculator, we find that the value of θ which satisfies this condition is approximately 9.42 radians. Since cos(θ) > 0 in this range, and cos(θ) > 0 when 8π≤θ≤10π, this value of θ satisfies all the given conditions.

Therefore, the angle θ so that the beetle is exactly at the surface of the water is approximately 9.42 radians.