Where Space Curve Intersects XZ Plane? C=r(t)=<2-(t^2), 2t-1, ln t)

crybloodwing

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Aug 22, 2017
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So I have a problem, and I believe I got the right answer, however the teacher says something different and I am not sure if I am wrong, or he wrote the answer wrong.

So Curve C=r(t)=<2-(t^2), 2t-1, ln t)

We want a point on C that intersects the XZ plane, so we want y=0.

so 2t-1=0
2t=1
t=.5

Then I put t back into the other points.

x=2-(.5^2)
x=2-.25
x=1.75

Teacher said X was 7/4, which is the same value as what I got.

Y still = 0

Z=ln(.5)

However, the teacher said z=ln(2), which is the same value, except it is positive, while .5 is negative.

My point: (1.75, 0, ln(.5))
Teacher's Point: (1.75, 0, ln(2))

Any help? Am I right and the teacher wrote the z-value wrong? Or did I miss a step or logic somewhere?
 
So I have a problem, and I believe I got the right answer, however the teacher says something different and I am not sure if I am wrong, or he wrote the answer wrong.

So Curve C=r(t)=<2-(t^2), 2t-1, ln t>

We want a point on C that intersects the XZ plane, so we want y=0.

so 2t-1=0
2t=1
t=.5

Then I put t back into the other points.

x=2-(.5^2)
x=2-.25
x=1.75

Teacher said X was 7/4, which is the same value as what I got.

Y still = 0

Z=ln(.5)

However, the teacher said z=ln(2), which is the same value, except it is positive, while .5 is negative.

My point: (1.75, 0, ln(.5))
Teacher's Point: (1.75, 0, ln(2))

Any help? Am I right and the teacher wrote the z-value wrong? Or did I miss a step or logic somewhere?

Assuming you copied the problem correctly, and it wasn't -ln(t), you are correct. It looks like the teacher lost a negative sign, and meant z = -ln(2).
 
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