Vector Valued Function and values of t parallel to the xy-plane

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SUMMARY

The discussion focuses on determining the values of t for which the derivative of the vector-valued function r(t) = (√(t+1), cos(t), t^4 - 8t^2) is parallel to the xy-plane. The key insight is that for r'(t) to be parallel to the xy-plane, the z-component of r'(t) must equal zero. The derivative is calculated as r'(t) = (1/2(t+1)^(-1/2), -sin(t), 4t^3 - 16t), leading to the condition 4t^3 - 16t = 0 for finding the relevant values of t.

PREREQUISITES
  • Understanding of vector-valued functions
  • Knowledge of derivatives and their applications
  • Familiarity with the concept of parallel vectors
  • Ability to perform cross-products of vectors
NEXT STEPS
  • Study the implications of setting the z-component of a vector to zero in vector calculus
  • Learn how to solve polynomial equations, specifically cubic equations like 4t^3 - 16t = 0
  • Explore the geometric interpretation of vector derivatives in three-dimensional space
  • Investigate the properties of cross-products in determining vector relationships
USEFUL FOR

Students studying calculus, particularly those focusing on vector calculus, as well as educators looking for examples of vector-valued functions and their derivatives.

Starlit_day
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Homework Statement


So, the problem is this:
Find all values of t such that r'(t) is parallel to the xy-plane.
And my equation is:
r(t)=(Squareroot(t+1) , cos(t), t4-8t2)


Homework Equations



Well, I will definitely have to know how to take the dirivative of the given vector valued function, which I know how to do. I think An important thing to know would be: the cross-product of the two vector-valued functions. But I can't really think of any pertinant formulae off hand that would help with this problem ... (maybe that's why I'm confused).

The Attempt at a Solution


Okay, so as I said I need to take the derivative and then I thought take the cross-product of the derivative and the original function.
So I did that and got: r'(t)= (1/2(t+1)-1/2, -sin(t), 4t3- 16t)
Wasn't much of a problem there, but now I'm kinda clueless as to how I would find ALL values of t for r'(t) parallel to the xy-plane. So I was thinking about the cross-product because that will tell me if the vector valued function and its derivative are parallel (if I get 0 as my answer when I do the cross-product) then maybe I can figure out a parametric equation for a plane that contains the vector function (the first one not the derivative), and all values in that plane would be parallel to the derivative because the plane is parallel to the derivative. >.< That sounds kind of complicated and somewhat off base to me. I'm speculating here because I really am not sure what to do. Although, that really wouldn't help me much considering I have to deal with the xy-plane. Hm,
So does anyone think they could help steer me in the right direction conceptually? I think I'm stuck conceptually on how I would go about finding these values of t ... I mean I know what they mean by a Vector Valued Function I understand what a function is and what a vector is I know about derivatives and understand all of that and I know when two vectors are parallel when given the corrodinates, but I just don't get this problem. Could someone help me at least start thinking in the right direction.
2
 
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Starlit_day said:

Homework Statement


So, the problem is this:
Find all values of t such that r'(t) is parallel to the xy-plane.
And my equation is:
r(t)=(Squareroot(t+1) , cos(t), t4-8t2)


Homework Equations



Well, I will definitely have to know how to take the dirivative of the given vector valued function, which I know how to do. I think An important thing to know would be: the cross-product of the two vector-valued functions. But I can't really think of any pertinant formulae off hand that would help with this problem ... (maybe that's why I'm confused).

The Attempt at a Solution


Okay, so as I said I need to take the derivative and then I thought take the cross-product of the derivative and the original function.
So I did that and got: r'(t)= (1/2(t+1)-1/2, -sin(t), 4t3- 16t)
Wasn't much of a problem there, but now I'm kinda clueless as to how I would find ALL values of t for r'(t) parallel to the xy-plane. So I was thinking about the cross-product because that will tell me if the vector valued function and its derivative are parallel (if I get 0 as my answer when I do the cross-product) then maybe I can figure out a parametric equation for a plane that contains the vector function (the first one not the derivative), and all values in that plane would be parallel to the derivative because the plane is parallel to the derivative. >.< That sounds kind of complicated and somewhat off base to me. I'm speculating here because I really am not sure what to do. Although, that really wouldn't help me much considering I have to deal with the xy-plane. Hm,
So does anyone think they could help steer me in the right direction conceptually? I think I'm stuck conceptually on how I would go about finding these values of t ... I mean I know what they mean by a Vector Valued Function I understand what a function is and what a vector is I know about derivatives and understand all of that and I know when two vectors are parallel when given the corrodinates, but I just don't get this problem. Could someone help me at least start thinking in the right direction.
2

For r'(t) to be parallel to the x-y plane, its z component has to be zero.
 

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