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Derivatives and Vectors

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Homework Statement


A point particle has a position vector r⃗ (t) as a function of time t, given by

r(t)=(1−t2)x^−2t(t+5)y^+8(t+2)z^.

where distances are in meters, and time t is in seconds. Now, let t=t1= 23 s.

What is the (smaller) angle between the velocity vector at time t1 and the z^ axis? (in degrees)

Homework Equations





The Attempt at a Solution


I took the derivative of the position vector to get -2tx-(4t+10)y+8z. At t=23 the y component of the velocity is -102, but I don't know what to do from there.
 
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Answers and Replies

  • #2
UltrafastPED
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You can determine the angle by taking the dot product of the velocity vector with the given axis, then apply the relation that A .dot. B = |A|x|B| cos(angle between A and B).
 
  • #3
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Is the answer 93.22 degrees?
 
  • #4
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Does anyone know?
 
  • #5
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Anyone?
 
  • #6
Ibix
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I don't get that. How did you get it?

Note that your velocity vector's y component is not consistent with the y component of r. Which one is correct?

Also, read the question carefully. You will see that your answer is obviously wrong.
 
  • #7
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Oops, I made a typo, but in my calculations I made the derivative of the y-component of the postion vector -4t-10, which should be right. According to the dot product A_x*B_x+A_y*B_y+A_z*B_z=A*B*cos(alpha). B_x and B_y =0 so you are left with A_z*B_z=A*B*cos(alpha) after you substitute the vectors you get 8z=(-2t-4t-10+8)*z*cos (alpha). Simplifying you get 8=(-6t-2)*cos(alpha). Since t=23, 8=-140*cos(alpha), and cos(alpha)=8/-140 which equals 93.22 degrees. What am I doing wrong?
 
  • #8
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Oops, I made a typo, but in my calculations I made the derivative of the y-component of the postion vector -4t-10, which should be right. According to the dot product A_x*B_x+A_y*B_y+A_z*B_z=A*B*cos(alpha). B_x and B_y =0 so you are left with A_z*B_z=A*B*cos(alpha)
Your problem comes when you "substitute the vectors". In the equation ##\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| cos(\theta)##, ##|\vec{A}|## and ##|\vec{B}|## refer to the magnitudes of ##\vec{A}## and ##\vec{B}##, not the equations of the vectors themselves. Do you know what the equation for the magnitude of a vector is?
 
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  • #9
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It's the square root of the respective components squared, right?
 
  • #10
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It's the square root of the respective components squared, right?
Right! So what's the magnitude of the velocity vector at the specified time, and what's the magnitude of the vector ##\hat{z}##?

Oh, and another error comes in when you take the dot product. You have the definition of a dot product right, but remember that you should get a scalar (just a number) as a result. But what you have is ##8\vec{z}##, which is a vector. Do you see how to fix that?
 
  • #11
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Ok I redid my calculations and got alpha=85.91 degrees. How's that?
 
  • #12
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Ok I redid my calculations and got alpha=85.91 degrees. How's that?
Exactly, that's what I got too!
 
  • #13
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Just to make sure, is the magnitude of the z component of vector B just 1?
 
  • #14
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Just to make sure, is the magnitude of the z component of vector B just 1?
Yep! You could calculate its magnitude as you would any other vector or use the fact that it's a unit vector, which by definition has a magnitude of 1, but either way that's right!
 
  • #15
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What is a unit vector?
 
  • #16
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What is a unit vector?
A unit vector is any vector whose magnitude is one (or unity). That's the mathematical definition, but I'll explain what it is in the context of the problem as well.

You already know that you can take a vector ##\vec{A}## and decompose it into its three components ##A_x##, ##A_y##, and ##A_z## right? But you can't just say that $$A_x + A_y + A_z = \vec{A}$$, because then you have a scalar quantity on the left hand side of the equation and a vector quantity on the right, which doesn't make any sense. So that equation is wrong.

The proper equation is $$A_x \hat{x} + A_y \hat{y} + A_z \hat{z} = \vec{A}$$ ##\hat{x}##, ##\hat{y}##, and ##\hat{z}## are unit vectors, which means they all have a magnitude of one. So the way this equation works is that if you start at the origin, and then you add the vector ##A_x \hat{x}##, and then you add the vector ##A_y \hat{y}##, and then you add the vector ##A_z \hat{z}##, you get ##\vec{A}##! Does that much make sense?

##A_x \hat{x}##, ##A_y \hat{y}##, and ##A_z \hat{z}## are the component vectors of ##\vec{A}##, and ##A_x##, ##A_y##, and ##A_z## are the magnitudes of these component vectors. This is because the magnitude of the unit vectors is one, so when you multiply any number by them, the magnitude of the resulting vector is just the magnitude of the number. You can actually derive the equation for the magnitude of a vector from the above equation!

So what happened when you took the dot product of ##\vec{A}## and ##\vec{B}## is you actually performed $$A_x B_x \hat{x} \cdot \hat{x} + A_y B_y \hat{y} \cdot \hat{y} + A_z B_z \hat{z} \cdot \hat{z}$$ And then, since you know that ##\hat{x} \cdot \hat{x} = |\hat{x}| |\hat{x}| cos(\theta)##, and you know that the magnitude of ##\hat{x}## is 1 and the angle between them is 0 (because they're the same vector), then that just turns to 1! So your equation goes to ##A_x B_x + A_y B_y + A_z B_z##, which is what you're probably used to.
 
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  • #17
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How do you derive the equation for the magnitude of a vector? And are there 2 x,y,and z hats because each vector provides an x,y,and z hat?
 
  • #18
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How do you derive the equation for the magnitude of a vector? And are there 2 x,y,and z hats because each vector provides an x,y,and z hat?
That's exactly why there are two of them! Sorry for not being explicit about that.

So we have a vector ##\vec{A}##, right? Let's just take the dot product ##\vec{A} \cdot \vec{A}##. Not for any particular reason, let's just do it. We get $$\vec{A} \cdot \vec{A} = A_x^2 + A_y^2 + A_z^2$$ Makes sense, right? But we also know that $$\vec{A} \cdot \vec{A} = |\vec{A}||\vec{A}|cos(\theta)$$ So we can just equate these two and we get $$|\vec{A}||\vec{A}| cos(\theta) = A_x^2 + A_y^2 + A_z^2$$ We know that the angle between ##\vec{A}## and ##\vec{A}## is zero, so we get $$|\vec{A}|^2 = A_x^2 + A_y^2 + A_z^2$$ Taking the square root of each side, we arrive at the familiar equation $$|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}$$
 
  • #19
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Okay, thanks for all your help!
 

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