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At what points does the curve r(t) = ti + (3t − t2)k intersect the paraboloid z = x2 + y2? (if an answer does not exist, enter dne.) (x, y, z) = (smaller t-value) (x, y, z) = (larger t-value)

Respuesta :

LammettHash

[tex]\mathbf r(t)=x(t)\,\mathbf i+y(t)\,\mathbf j+z(t)\,\mathbf k=t\,\mathbf i+(3t-t^2)\,\mathbf k[/tex]


Along the curve [tex]\mathbf r(t)[/tex], we have [tex]y(t)=0[/tex], so the equation of the paraboloid reduces to


[tex]z=x^2\iff3t-t^2=t^2\implies2t^2-3t=t(2t-3)=0\implies t=0\text{ and }t=\dfrac32[/tex]

yoodyannapolis

We have given curve and paraboloid as

[tex]r(t) = ti + (3t-t^{2} )k[/tex],[tex]z=x^{2} +y^{2}[/tex]

And along the given curve [tex]r(t)[/tex], we have [tex]y(t)=0[/tex], so the equation of the paraboloid is

[tex]z=x^{2}[/tex] ⇔[tex]3t^{2} -t^{2} =t^{2}\\[/tex]

⇒

⇒[tex]t(2t-3)=0[/tex]

⇒[tex]t=0,t=\frac{3}{2}[/tex]

Therefore at the point [tex](\frac{3}{2} ,0,\frac{3}{2} )[/tex] the curve [tex]r(t) = ti + (3t-t^{2} )k[/tex] intersect the paraboloid [tex]z = x^{2} + y^{2}[/tex].

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