2. Let 7(t) = (t°,tv2, In(t)). (a) Find 7"(t). rc)=くt,2,> (b) Find T(t) and simplify completely. = Nt +2 + <ち, TC6) =' Tr'et) (c) Find the tangent line to the curve generate by 7 at the point (, /2,0). ちもtE, nce): 0 %3D =>t=さしt=1,t= Means for polnt c, NE,0), t=1 at t:1 then Means tangent lines divectonal fatio of curve at Point <5, Nž,07 will be <, NZ,1>
2. Let 7(t) = (t°,tv2, In(t)). (a) Find 7"(t). rc)=くt,2,> (b) Find T(t) and simplify completely. = Nt +2 + <ち, TC6) =' Tr'et) (c) Find the tangent line to the curve generate by 7 at the point (, /2,0). ちもtE, nce): 0 %3D =>t=さしt=1,t= Means for polnt c, NE,0), t=1 at t:1 then Means tangent lines divectonal fatio of curve at Point <5, Nž,07 will be <, NZ,1>
2. Let 7(t) = (t°,tv2, In(t)). (a) Find 7"(t). rc)=くt,2,> (b) Find T(t) and simplify completely. = Nt +2 + <ち, TC6) =' Tr'et) (c) Find the tangent line to the curve generate by 7 at the point (, /2,0). ちもtE, nce): 0 %3D =>t=さしt=1,t= Means for polnt c, NE,0), t=1 at t:1 then Means tangent lines divectonal fatio of curve at Point <5, Nž,07 will be <, NZ,1>
Vector Value Functions! I tried solving this on my own and supposedly there's something wrong with the way answer these questions. Is there any way you could identify where I went wrong and what I need to do to fix it?
Transcribed Image Text:2. Let 7(t) = (²,tv2, In(t)).
(a) Find "(t).
riegsくt,E,>
(b) Find T(t) and simplify completely.
TC4) =Triet)
ノ
(c) Find the tangent line to the curve generate by 7 at the point G, /2,0).
ちセEニ, Ance): 0
=>t=さしt=1しセ=
Means for polat cE, NE,O), t=1
at t:1 then
Means tangent lines divectonal
fatio of curve at Point
<z, NZ,0> will be<', NZ,1?
so
tangent
line eauation 's
rce)-く支,6>+t<\,WE,>
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
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