Vectors Calaulus 1.7

Velocity and Acceleration   Definition of Velocity and Speed In single variable calculus the velocity is defined as the derivative of the position function.  For vector calculus, we make the same definition.   Definition of Velocity Let r(t) be a differentiable vector valued function representing the  position vector of a particle at time t.  Then the velocity vector is the derivative of the position vector.                v(t)  =  r'(t)  =  x'(t)i + y'(t)j + z'(t)k   Example Find the velocity vector v(t) if the position vector is          r(t)  =  3ti + 2t2j - sin t k  Solution We just take the derivative         v(t)  =  3i + 4tj + cos t k  When we think of speed, we think of how fast we are going.  Speed should not be negative.   In one variable calculus, speed was the absolute value of the velocity.  For vector calculus, it is the magnitude of the velocity.   Definition of Speed Let r(t) be a differentiable vector valued function representing the position  of a particle.  Then the speed of the particle is the magnitude of the  velocity vector.           Speed  =  || v(t) ||  =  || r'(t) ||   Example Let          r(t)  = 3i + 2t j + cos t k Find the speed after p/4 seconds.   Solution We first find the velocity vector         v(t)  =  r'(t)  =  2 j - sin t k We have          v(p/4)  =  2 j - /2 k Its magnitude is the square root of the sum of the squares or         Speed  =  || v ||  =    Acceleration In one variable calculus, we defined the acceleration of a particle as the second derivative of the position function.  Nothing changes for vector calculus.   Definition of Acceleration Let r(t) be a twice differentiable vector valued function representing the  position vector of a particle at time t.  Then the acceleration vector is the second derivative of the position vector.                a(t)  =  r''(t)  =  x''(t)i + y''(t)j + z''(t)k   Example Find the velocity and acceleration of the position function         r(t)  = 4t i + t2 j  when t  =  -1.  Then sketch the vectors. Solution  The velocity vector is         v(t)  =  r'(t)  =  4 i  + 2t j Plugging in -1 for t gives         v(-1)  =  4 i - 2j Take another derivative to find the acceleration.         a(t)  =  v'(t)  =  2j Below is a picture of the vectors.           Projectile Motion Since the velocity and acceleration vectors are defined as first and second derivatives of  the position vector, we can get back to the position vector by integrating. Example You are a anti-missile operator and have spotted a missile heading towards you at the position         re  =  1000i + 500j with velocity            ve  =  -30i + 3j  You can fire your anti-missile at 100 meters per second.  At what angle should you fire it  so that you intercept the missile.  Assume that gravity is the only force acting on the projectiles. Solution The acceleration vector of the enemy missile is          ae(t)  =  -9.8 j  Integrating, we get the velocity vector         ve(t)  =  v1 i + (v2 - 9.8t) j Setting t  =  0 and using the initial velocity of the enemy missile gives         ve(t)  =  -30 i + (3 - 9.8t) j Now integrate again to find the position function         re(t)  =  (-30t + r1) i + (-4.9t2 + 3t + r2) j Again setting t  =  0 and using the initial conditions gives         re(t)  =  (-30t + 1000) i + (-4.9t2 + 3t + 500) j The acceleration of your anti-missile-missile is also         ay(t)  =  -9.8 j  Integrating, we get the velocity vector         vy(t)  =  v1 i + (v2 - 9.8t) j Since the magnitude of our velocity is 100, we can say         vy(0)  =  100 cos q i + 100 sin q j So that         vy(t)  =  100 cos q i + (100 sin q - 9.8t) j Now integrate again to find the position function         ry(t)  =  (100t cos q + r1) i + (-4.9t2 + 100t sin q + r2) j Our anti-missile-missile starts out at base, so the initial position is the origin.  All the  constants are zero.         ry(t)  =  (100t cos q) i + (-4.9t2 + 100t sin q) j Since we want to intercept the enemy missile, we set the position vectors equal to each other.         (100t cos q) i + (-4.9t2 + 100t sin q) j  =  (-30t + 1000) i + (-4.9t2 + 3t + 500) j Equating coefficients gives         100t cos q =  -30t + 1000         -4.9t2 + 100t sin q  =  -4.9t2 + 3t + 500 The first equation gives                           1000         t  =                                                    100cos q  +  30 Simplifying the second equation and substituting gives              100000 sin q                    3000                                       =                                  +  500                    100cos q  +  30          100cos q +  30         Clear denominators to get         100000 sin q  =  3000 + 50000 cos q + 15000 At this point we use a calculator to solve for q to         q  =  .62535 radians