- #1
SlideMan
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Just wanted to check and see if this is right. The k-component of the vector is what I'm unsure of...I've always sucked at converting to parametric form. :)
Convert to parametric form:
x^{2} + y^{2} = 9, z = 4arctan(y/x)
The i- and j-components of the vector are obviously 3cos(t) and 3sin(t), respectively. I'm not sure how the k-component is supposed to turn out...
Here's my attempt:
x = 3cos(t)
y = 3sin(t)
So, z = 4 arctan \frac{3sin(t)}{3cos(t)} = 4 arctan(tan(t)) = 4t
Thus, r(t) = [3cos(t), 3sin(t), 4t]
Homework Statement
Convert to parametric form:
x^{2} + y^{2} = 9, z = 4arctan(y/x)
The Attempt at a Solution
The i- and j-components of the vector are obviously 3cos(t) and 3sin(t), respectively. I'm not sure how the k-component is supposed to turn out...
Here's my attempt:
x = 3cos(t)
y = 3sin(t)
So, z = 4 arctan \frac{3sin(t)}{3cos(t)} = 4 arctan(tan(t)) = 4t
Thus, r(t) = [3cos(t), 3sin(t), 4t]