A searchlgith on a ship 1km from shore casts a spot of lgiht on the shore line. The searchlight is rotating at a rate of 1 rpm. How fast is the spot on the shore mvoing at the point nearest the searchlight? How fast is ti moving when the spot is 250 m father down the shore?
(i'm allowed to use the TI-83 after finding the equation for the answers...)
plz show me how u got the answers plz!!! beggin ya
2007-11-12 14:11:59 · 2 answers · asked by SDGD 1 in Science & Mathematics ➔ Mathematics
Draw a right triangle. One leg, of length 1 (km), connects the ship to the spot on the shore nearest the ship (and the searchlight). Let the length (in km) of the other leg, which lies along the shore, be denoted x. The hypotenuse of this triangle coincides with the searchlight beam. Finally, let the angle opposite x, and adjacent to the leg of length 1, be θ. Convince yourself that dθ/dt is the rotation rate of the searchlight, and that dx/dt is the rate at which the spot of light moves along the shoreline. From the triangle,
tan(θ) = opposite/adjacent = x/1 = x so
θ = tan^(-1) (x) (that is, inverse tangent of x)
Differentiate both sides with respect to t, then use that result to evaluate dx/dt for the two requested values of x. Don't forget to change dθ/dt to radians/minute.
2007-11-12 14:31:51 · answer #1 · answered by Ron W 7 · 0⤊ 0⤋
initially - confirm to work out any commonalities in it working example on #a million all of it's divided by making use of 5x - so pull that out first 5x (x^2 - 2x -8) then you definately can element the parenthetical expression 5x (x - 4)(x - 2) try doing that first - then see if it helps slightly
2016-11-11 07:58:19 · answer #2 · answered by ? 4 · 0⤊ 0⤋