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Sketch the curve with the given vector equation. Indicate with an arrow the direction in which t increases. r(t) = 4 cos(t)i + 4 sin(t)j + "k"

Sagot :

Answer:

The equation to this question is "[tex]x^2+y^2=16[/tex]"

Step-by-step explanation:

Given value:

[tex]\to r(t) = 4 \cos (t)i + 4 sin(t)j + k\\where \\\\\to x(t)= 4 \cos t\\\to y(t) = 4 \sin t\\\to z(t)=1\\[/tex]

[tex]\to x^2+y^2= 16 \cos^2 t +16 \sin^2 t\\\\\to x^2+y^2= 16 (\cos^2 t +\sin^2 t)\\\\\therefore (\cos^2 t +\sin^2 t =1)\\\\\to x^2+y^2= 16 (1)\\\\\to x^2+y^2= 16\\\\\to z=1[/tex]

 So, the curve [tex]r(t) = 4 \cos (t)i + 4 sin(t)j + k[/tex] is basically a circle, which is defined in the attachment file.

View image Codiepienagoya
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