<p>I just bumped into this problem in calculus and it left me somewhat perplexed. I would appreciate any help on the problem.</p>
<p>The problem states, the velocity vector of an object moving in 2-space is expressed by <ln(t^2+2t),2t^2>. Find the acceleration vector of the object at t=2.</ln(t^2+2t),2t^2></p>
<p>My first reaction was to normalize the vector to find the direction of the velocity. I remember from physics that the acceleration should be the same direction as the velocity. Then I would multiply the normalized vector by the length of dy/dx, using the standard method for finding the tangent line to parametric equations.</p>
<p>However, my teacher provided the much simpler solution: simply differentiate the parametric equations to get the acceleration vector, <2t+2/t^2+2t, 4t>. Plugging in 2 for t and the answer becomes <3/4, 8> which is the answer.</p>
<p>After thinking about it for a while, I found that both approaches make sense. But they definitely don't provide the same answer, and the second approach does not produce an acceleration vector in the same direction as the velocity vector. I could find no way to reconcile the two solutions. Please help.</p>
<p>you wouldn't have to account for the velocity vector, because as your teacher was correct in his assumption, it's just the change in velocity, it doesn't matter what the bearing is... therefore, it is clear that the acceleration is just taking into account positive values of acceleration... which makes sense since t=2>0</p>
<p>i agree with your teacher... simply because the rate of change in the y plane and x plane is relative to the acceleration in the x and y plane... therefore...<3/4, 8> is the answer... i just thought it worked because it was for positive values... but, i guess it would work for all values... sorry, i have difficulty putting my math thoughts into words</p>
<p>That makes sense. My original problem with that answer was this: the velocity vector at t=2 is <ln(8), 8="">. That means the velocity vector is not in the same direction as the acceleration vector. But now that I think about it, the two really isn't in the same direction in this case. The object must be moving in one direction but accelerating in another. So <3/4,8> is the answer. Thanks.</ln(8),></p>
<p>I remember from physics that the acceleration should be the same direction as the velocity.==>
err! wrong concept here!
acceleration should be the same direction as the CHANGE in the velocity .. not the velocity...e.g. centripetal acceleration is directed to the centre of the circle and is definitely of a different direction to the linear velocity vector at any position.</p>
<p>^^ yeah. Think about a ball you throw up into the air. Once it leaves your hand, it's accelerating down towards the earth, but still moving upwards.</p>
<p>for a lot of AP problems, they don't make you think too much, they just want you to demonstrate concepts. dx/dt = v dv/dt = a da/dt = jerk</p>
<p>Try it simply first, and if some problem crops up, then look for a clever version.</p>