Find rectangular coordinates for the point which has polar coordinates (6, pi/2)
We begin by recalling the relationships between the rectangular and polar coordinates. --> x = r*cos(theta) and y = r*sin(theta) where theta is the angle and r is the radius --> rectangular coordinates are represented as ( x , y ) on graphs and polar coordinates are represented as (r, theta ) We find the rectangular coordinates as follows: --> x = r*cos(theta) = 6*cos(pi/2) = 0 --> y = r*sin(theta) = 6*sin(pi/2) = 6 Ans: (0 , 6)
What is the period of the trigonometric function given by f(x) = 2 sin(5x)
We first begin by noticing that this is a sine function. Sine functions are called periodic functions meaning that they will repeat forever. The general equation for the sine function is f(x) = Asin(Bx+C) + D where A represents the amplitude, B represents the angular frequency, C represents the phase shift, and D represents the vertical shift. The period a function is given by (2*pi/B). In our case, the B = 5; therefore, the period is (2*pi/5)
Particle Motion: Displacement of a particle is given by s (t) = 5t^2-3t+7 where t is time. 1) Find Velocity v(t) 2) At what time does v(t)=0 (What does this mean?) 3) Find Acceleration a(t)
Begin by remembering the relationship between displacement, velocity, and acceleration. Notice that the derivative of displacement is velocity and the derivative of velocity is acceleration. 1) Find Velocity v(t). Simply take the derivative of s(t). s'(t) = 10t-3 = v(t). The answer is v(t) = 10t-3 2) At what time does v(t) = 0. Graphically, we know that this is a critical point of s(t) meaning that we have reached either a maximum or minimum i.e. the derivative (aka slope) at this point is 0. From the equation described by s(t), we know that it's a parabola and a parabola has a critical point. From the point of view of the particle, it is switching direction at this point. --> v(t) = 0 = 10t-3 occurs at t=3/10 3) Find acceleration a(t). Simply take the derivative of v(t). v'(t) = 10. The answer is a(t) = 10.