Answer:
t = -2,2
Step-by-step explanation:
The curve equation, [tex]r(t)=2ti+t^2j-t^2k[/tex]
Let at t=a tangent to the curve.
Equation of tangent:
L(k) = r(a) + k r'(a)
[tex]r(a)=<2a,a^2,-a^2>[/tex]
[tex]r'(t)=<2,2t,-2t>[/tex]
Put t=a
[tex]r'(a)=<2,2a,-2a>[/tex]
Equation of tangent:
[tex]L(k) = <2a,a^2,-a^2> + k<2,2a,-2a>[/tex]
[tex]L(k)=<2a+2k,a^2+2ak,-a^2-2ak>[/tex]
Tangent passing through (0,-4,4)
Therefore,
2a + 2k = 0 ⇒ k = -a
[tex]a^2+2ak=-4[/tex]
[tex]a^2-2a^2=-4[/tex]
[tex]-a^2=-4[/tex]
[tex]a=\pm 2[/tex]
At t = -2,2 tangent line to the curve passing through (0,-4,4)
