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Derivatives of coordinate equations

  1. Dec 14, 2013 #1
    1. The problem statement, all variables and given/known data
    We have two coordinate functions of time, as follows: x(t) = 5 + 2t ; y(t) = -3+3t+2t2. Find velocity [itex]\vec{v}[/itex], acceleration [itex]\vec{a}[/itex], tangential acceleration [itex]\vec{a_t}[/itex], normal acceleration [itex]\vec{a_n}[/itex] functions of time and their magnitude's functions of time.

    2. Relevant equations
    [itex]\frac{dx}{dt} = v[/itex]
    [itex]\frac{dv}{dt} = a[/itex]

    3. The attempt at a solution
    So I guess [itex]v_x = \frac{dx(t)}{dt} = 2[/itex] and [itex]v_y = \frac{dy(t)}{dt} = 3+4t[/itex]. And so [itex]\vec{v_x} = 2\vec{i_y}[/itex]; [itex]\vec{v_y} = (3+4t)\vec{i_y}[/itex]. I guess that [itex]v = \sqrt{v_x^2 + v_y^2} = \sqrt{13+24t+16t^2}[/itex]. But then how do you find the vector of v? And also, am I right saying that at is derivative of vx and an is derivative of vy?
  2. jcsd
  3. Dec 14, 2013 #2
    The velocity vector [itex]\vec{v}[/itex] is just the sum of the x and y vectorial components. Incidentally, your equation [itex]\vec{v_x} = 2\vec{i_y}[/itex] has a typo. It should read [itex]\vec{v_x} = 2\vec{i_x}[/itex]. The acceleration vector is just [itex]\vec{a}=\frac{d\vec{v}}{dt}[/itex]. Just take the time derivative of [itex]\vec{v}[/itex]. The tangential acceleration is just the component of [itex]\vec{a}[/itex] in the same direction as the velocity vector. The normal acceleration is the component of [itex]\vec{a}[/itex] in the direction perpendicular to the velocity vector. Do you know how to determine the component of [itex]\vec{a}[/itex] in the same direction as the velocity vector? Hint: You might start out by finding the equation for a unit vector in the same direction as the velocity vector.


  4. Dec 14, 2013 #3
    So [itex] \vec{v} = \vec{v_x}+\vec{v_y} = 2\vec{i_x}+(3+4t)\vec{i_y}[/itex], and then [itex] \vec{a} = \frac{d\vec{v}}{dt} = 4\vec{i_y}[/itex]? Am I right? And how do I derive the magnitude of [itex]\vec{a}[/itex]? Is this equation [itex]a=\sqrt{(\frac{dv_x}{dt})^2 + (\frac{dv_y}{dt})^2}[/itex] correct (from such calculations I get a = 4)?
    Now for the unit vector of in the same direction as velocity vector, I guess the equation looks something like this: [itex]\vec{i_t} = \frac{\vec{v}}{v} = \frac{2\vec{i_x}+(3+4t)\vec{i_y}}{\sqrt{13+24t+16t^2}}[/itex]. Is it correct to state that [itex]\vec{a_t} = a * \vec{i_t}[/itex] then?
  5. Dec 14, 2013 #4
    Yes. This is the correct result for the unit vector. To get the tangent component of the acceleration, you need to take the dot product of the acceleration vector with the unit vector in the direction of the velocity. Once you do this, can you think of how to get the normal component of the acceleration?
  6. Dec 14, 2013 #5
    So [itex]a_t = \vec{a} * \vec{i_t} = a_y i_{ty} + a_x i_{tx} = 4*\frac{3+4t}{\sqrt{13+24t+16t^2}} + 0*\frac{2}{\sqrt{13+24t+16t^2}} = \frac{12+16t}{\sqrt{13+24t+16t^2}}[/itex]? And it seems to me that [itex]a=\sqrt{a_t^2+a_n^2}[/itex] and so [itex]a_n=\sqrt{a^2-a_t^2}=\sqrt{16-(\frac{12+16t}{\sqrt{13+24t+16t^2}})^2} =\sqrt{\frac{64}{13+24t+16t^2}}[/itex]. And probably [itex]\vec{a_t} = a_t*\vec{i_t}[/itex]? But how do I find unit vector perpendicular to velocity vector?
  7. Dec 14, 2013 #6
    Very, very, very nicely done. Now, if [itex]\vec{a_t} = a_t*\vec{i_t}[/itex], and you know the overall acceleration vector, then the normal component of the acceleration vector must be whatever is left over.
  8. Dec 14, 2013 #7
    Oh, that's right :) thank you for the big help, now all seems not so complicated afterall!
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