Consider the vector function r(t) =< 4t + 1, t2 − 3t, cos(t) >. (a) Find the unit tangent vector T at (1, 0, 1) I know that T = r'(t)/|r-(t)| and r'(t) = < 4, 2t-3, -sin(t)> I'm not sure why the book plugs in 0 for t. Thanks in advance
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Originally Posted by khuezy Consider the vector function r(t) =< 4t + 1, t2 − 3t, cos(t) >. (a) Find the unit tangent vector T at (1, 0, 1) I know that T = r'(t)/|r-(t)| and r'(t) = < 4, 2t-3, -sin(t)> I'm not sure why the book plugs in 0 for t. Thanks in advance Ok So as you noted So we can compute this to be So
Last edited by Mathstud28; Nov 14th 2008 at 05:35 PM. Reason: typo
In the answer key, they did this T(0) to get <4, -3, 0> as r'(0) Why did they use 0?
Originally Posted by khuezy In the answer key, they did this T(0) to get <4, -3, 0> as r'(0) Why did they use 0? at the point implies that there is a t value such that and It should be a little clearer why they picked ... --Chris
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