Find parametric equations for the tangent line to the curve

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The discussion centers on finding the parametric equations for the tangent line to a curve at the point (5, 2, 0). The user initially calculated the derivatives of the vector components but struggled to identify the correct value of t that corresponds to the given point. After some confusion, it was clarified that t should be set to 1 to accurately reflect the point on the curve. The user ultimately derived the correct equations, concluding with x(t) = 5 + 2t and z(t) = 2t. This highlights the importance of correctly determining the parameter t when working with parametric equations.
Atlos
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Homework Statement



symimage.gif

Homework Equations



r = <x,y,z>
r' = <x',y',z'>

The Attempt at a Solution



I started by finding the derivatives of each part of the vector and got:

x= 2/sqrt(t) y= 3t^2+1 z= 3t^2-1

Then I plugged the point (5,2,0) into that and got (2/sqrt(5), 13, -1). This should be multiplied by t. Then by plugging in (5,2,0) back into the original line, you get (1+4sqrt(5), 10, 0). I thought that the parametric equation for this would be (1+4sqrt(5), 10, 0) + (2/sqrt(5), 13, -1)*t, but it's not working. The solution y(t) = 2+4t is already given, so I'm obviously already wrong. Any idea what I did wrong?
 
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Atlos said:

Homework Statement



symimage.gif



Homework Equations



r = <x,y,z>
r' = <x',y',z'>

The Attempt at a Solution



I started by finding the derivatives of each part of the vector and got:

x= 2/sqrt(t) y= 3t^2+1 z= 3t^2-1
These would be x' = 2/sqrt(t), y'= 3t^2+1, z'= 3t^2-1

Atlos said:
Then I plugged the point (5,2,0) into that and got (2/sqrt(5), 13, -1).
Each of these derivatives is a function of t. At the point (5, 2, 0), what's the value of t?
Atlos said:
This should be multiplied by t. Then by plugging in (5,2,0) back into the original line, you get (1+4sqrt(5), 10, 0). I thought that the parametric equation for this would be (1+4sqrt(5), 10, 0) + (2/sqrt(5), 13, -1)*t, but it's not working. The solution y(t) = 2+4t is already given, so I'm obviously already wrong. Any idea what I did wrong?
 
Mark44 said:
These would be x' = 2/sqrt(t), y'= 3t^2+1, z'= 3t^2-1

Each of these derivatives is a function of t. At the point (5, 2, 0), what's the value of t?

Woops, I forgot the prime for those. What do you mean what would the value of t be? Like solve for 2/sqrt(t) = 5, 3t^2+1 = 2, and 3t^2-1 = 0 to find a t vector? Otherwise I don't see a value of t that would make the equations go through the point (5,2,0).

edit: oh nevermind t would be 1 to get to the point (5,2,0). So should I plug 1 into the derivative for each?
 
Solve for t in the equations for x, y, and z, at the point (5, 2, 0), NOT the equations for x', y', and z'. (5, 2, 0) is a point on the curve described by the parametric equations.
 
Ah, got it! I understand it now, I got x(t)=5+2t and z(t)=2t which were both right.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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