Given the curve $$r(t)=(t,t^2,2)$$ I have to find the tangent vector to $r$ at $Q(1,1,2)$. From the coordinates of $Q$, I know that $t=1$, so the tangent vector is $$r'(1)=(1,2,0)$$ But when I plot the curve $r$ and the vector $r'(1)-r(1)=(0,1,-2)$ in Geogebra it's not tangent at all. What's wrong?
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$(1, 1, 2) + (1, 2, 0) s$ will be tangent to $r(t)$. – Math Lover Apr 24 '21 at 13:50
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@MathLover what is $s$? – mvfs314 Apr 24 '21 at 13:52
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a scalar, just like $t$ – Math Lover Apr 24 '21 at 13:53
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@MathLover So a tangent vector could be $(0,-1,2)$? – mvfs314 Apr 24 '21 at 13:56
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1Tangent vector at point $(1, 1, 2)$ is $(1, 2, 0)$ – Math Lover Apr 24 '21 at 13:58
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But tangent vector is just a direction. We need to also specify one of the points on the line in addition to tangent vector to uniquely identify the tangent line. – Math Lover Apr 24 '21 at 14:00
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@MathLover I need to find the directional derivative of a function in the direction of the tangent vector to $r$ at $Q$.So I thought to take $(1,2,0)$. – mvfs314 Apr 24 '21 at 14:01
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1please find the gradient of the function and do dot product with unit vector $\frac{1}{\sqrt5}(1, 2, 0)$. – Math Lover Apr 24 '21 at 14:03
1 Answers
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Note that you must plot the vector $(1,2,0)$ starting at $Q(1,1,2)$ (you have possibly drawn it starting from $(0,0,0)$). A sketch is as follows
Mostafa Ayaz
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