Solving Calculus Problems with Mechanics

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    Calculus Mechanics
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Discussion Overview

The discussion revolves around solving calculus problems related to mechanics, specifically focusing on finding tangent and normal lines for given paths in both two-dimensional and three-dimensional contexts. Participants explore the application of derivatives in these calculations and the relationship between velocity vectors and tangent lines.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions whether the derivative is sufficient for solving the problems related to tangent and normal lines.
  • Another participant states that the derivative provides a tangent vector, which can be used to find the equation of a line through a point in the direction of that vector.
  • A different participant breaks down the components of the path and calculates the velocity vector at a specific point, providing values for each direction.
  • There is uncertainty expressed regarding the calculation of the normal vector and whether it can be derived simply from the derivative.
  • One participant distinguishes between the velocity vector and the tangent line, explaining that the velocity vector includes both direction and magnitude, while the tangent vector represents direction only.

Areas of Agreement / Disagreement

Participants generally agree on the process of finding the tangent vector using derivatives, but there is uncertainty regarding the calculation of the normal vector and the distinction between the velocity vector and the tangent line. The discussion remains unresolved on these points.

Contextual Notes

Participants express varying levels of confidence in their understanding of the relationships between tangent lines, normal lines, and velocity vectors, indicating potential gaps in assumptions or definitions that are not fully explored.

Feynmanfan
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Hello everybody!

I'm having trouble with this calculus problem, where I don't know if I can apply what I've learned in MECHANICS.

"given a path s(t)=(t+1,E^t) calculate it's tangent line and the normal line at
this point s(0)"

and this is another version of the problem in R3

"given a path s(t)=(2t,t^2,Lnt) calculate the velocity vector and the tangent line at (2,1,0)"

How do I solve this? Is it just the derivative and that's all?
 
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the derivative gives a tangent vector at that point. you need to then finsd the equation of a line passing through that point in the direction of the tangent vector,
 
You can break it down like this:

If s(t) gives the path of the object, then its coordinates at all times are:
sx(t) = 2t
sy(t) = t2
sz(t) = ln(t)
Then the velocity in every direction (axis) is:
|vx(t)| = sx'(t) = 2
|vy(t)| = sy'(t) = 2t
|vz(t)| = sz'(t) = 1/t
For t = 1, when the object is at (2, 1, 0), you have |vx| = 2, |vy| = 4 and |vz| = 1. In other words, the velocity vector is 2i + 4j + k or (2, 4, 1).
 
thanks.
well I think I got that one but what about the normal vector? is it as easy as taking the derivative? is there a difference between the velocity vector and the tangent line?
 
Feynmanfan said:
well I think I got that one but what about the normal vector? is it as easy as taking the derivative?
The first problem you're not supposed to do with vectors, I don't think. Once you find the tangent line y = ax + b, the normal line is y = -x/a + c. You just need to find c...

Feynmanfan said:
is there a difference between the velocity vector and the tangent line?
The velocity vector determines the direction of the speed as well as its magnitude. Because of this you can't manipulate it however you want, because while the direction wouldn't change, the speed might. The tangent vector, on the other hand, only represents direction, which means its magnitude doesn't matter. Therefore you can multiply or divide it by any scalar k, i.e above we found the velocity vector to be (2, 4, 1) and that's it, but the tangent line can also be (4, 8, 2) or (1/2, 1, 1/4).
 
Last edited:
That was a great answer! Thanks a lot.
 

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