Vector valued functions: finding tangent line

In summary, the task at hand is to find the unit tangent vector T(t) and a set of parametric equations for the line tangent to the space curve at point P. To find T(t), the formula T(t) = r'(t)/||r'(t)|| is used, where r'(t) is the derivative of r(t). The point P is not directly used in finding T(t), but it is used to find the parametric equation of the line tangent to the space curve. The general form of a parametric equation of a line in vector form is used for this task.
  • #1
bfusco
128
1

Homework Statement


Find the unit tangent vector T(t) and find a set of parametric equations for the line tangent to the space curve at point P

r(t)= <2sin(t), 2cos(t), 4sin2(t)>, P(1, √3, 1)

The Attempt at a Solution


I found T(t) using the formula T(t)= r'(t)/||r'(t)||

r'(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>

∴ T(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>/ 2+8sin(t)cos(t)
-note: i could do more to the denominator as far as trig functions and algebra but i don't know if it is useful.

What i don't understand is how the point is used to find the parametric equation.
 
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  • #2
bfusco said:

Homework Statement


Find the unit tangent vector T(t) and find a set of parametric equations for the line tangent to the space curve at point P

r(t)= <2sin(t), 2cos(t), 4sin2(t)>, P(1, √3, 1)

The Attempt at a Solution


I found T(t) using the formula T(t)= r'(t)/||r'(t)||

r'(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>

∴ T(t)= <2cos(t), -2sin(t), 8sin(t)cos(t)>/ sqrt([STRIKE]2[/STRIKE] 4+8sin(t)cos(t))
-note: i could do more to the denominator as far as trig functions and algebra but i don't know if it is useful.

What i don't understand is how the point is used to find the parametric equation.
You have left out the square root for ||r' ||, as well as there is another error as noted above.

Do you know the general form of a parametric equation of a line in vector form?
 

Related to Vector valued functions: finding tangent line

1. What is a vector valued function?

A vector valued function is a mathematical function that maps a set of input values to a set of output values represented as vectors. It is denoted as f(t) = ⟨x(t), y(t), z(t)⟩, where x, y, and z are scalar functions of the independent variable t.

2. How do you find the derivative of a vector valued function?

To find the derivative of a vector valued function, you can use the chain rule. First, find the derivatives of the scalar functions x, y, and z with respect to t. Then, use these derivatives to calculate the derivative of the vector function f(t) = ⟨x(t), y(t), z(t)⟩ as f'(t) = ⟨x'(t), y'(t), z'(t)⟩.

3. What is the significance of the tangent line to a vector valued function?

The tangent line to a vector valued function represents the instantaneous rate of change of the function at a specific point. It can also be interpreted as the direction in which the function is changing at that point. This is useful in understanding the behavior of the function and making predictions about its future values.

4. How do you find the equation of the tangent line to a vector valued function?

To find the equation of the tangent line to a vector valued function at a specific point P, you need to find the slope of the tangent line and the coordinates of the point P. The slope can be calculated using the derivative of the vector function, and the coordinates of P can be found by plugging in the value of the independent variable at that point into the scalar functions x, y, and z. Finally, use the point-slope form of a line to write the equation of the tangent line.

5. How is the tangent line related to the graphical representation of a vector valued function?

The tangent line to a vector valued function at a specific point is the same as the tangent line to the curve representing the function on a graph at that point. This means that the slope of the tangent line can be visualized as the slope of the curve at that point, and the equation of the tangent line can be used to draw the tangent line on the graph.

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