Respuesta :
r(t)=(2t)i+(t²+2)j and r'(t)=2i+2tj.
r(t)=-ar'(t) where a is a scalar
2t=-2a and t²+2=-2at, equating vector components
So t=-a and a²+2=2a², from which a²=2 and a=±√2, making t=±√2.
r(±√2)=±2√2i+4j, (2√2,4) and (-2√2,4).
r(t)=-ar'(t) where a is a scalar
2t=-2a and t²+2=-2at, equating vector components
So t=-a and a²+2=2a², from which a²=2 and a=±√2, making t=±√2.
r(±√2)=±2√2i+4j, (2√2,4) and (-2√2,4).
The points on the curve r(t)=(2t)i+(t2+2)j at which r(t) and r′(t) have opposite direction is [tex](\pm\sqrt{2}, 4 )[/tex]
Given the following vector function
[tex]r(t)=(2t)i+(t^2+2)j[/tex]
If the points on the curve are in opposite directions then
[tex]r(t) = -ar'(t)[/tex]
Get the derivative of r(t)
[tex]r'(t) = 2i + 2tj[/tex]
The vector function becomes:
[tex](2t)i+(t^2+2)j = -a[2i + 2tj][/tex]
[tex]2ti+(t^2+2)j=-2ai-2atj[/tex]
compare the coefficients on both sides
[tex]2ti=-2ai\\t=-a\\[/tex]
Similarly:
[tex]t^2+2=-2at\\Since \ t = -a\\[/tex]
The equation becomes
[tex](-(-a))^2+2=-2a(-a)\\a^2+2= 2a^2\\\\2a^2-a^2=2\\a^2=2\\a=\pm\sqrt{2}[/tex]
Since [tex]t = -a[/tex]
[tex]t=-(\pm\sqrt{2} )\\t=\mp\sqrt{2}[/tex]
Substitute t = ±√2 into r(t)
Recall that
[tex]r(t)=(2t)i+(t^2+2)j\\r(t)=(2( \pm\sqrt{2}))i+(( \pm\sqrt{2})^2+2)j\\r(t) = \pm2\sqrt{2} + 4j\\[/tex]
From the given result we can see that the required point on the curve is [tex](\pm\sqrt{2}, 4 )[/tex]
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