I am currently working through questions in the textbook "Stewart Calculus: Early Transcendentals, 8th Edition". Fortunately, the textbook also has a solution manual but I'm having a hard time understanding why they did what they did.

The question is to find a vector equation for the tangent line to the curve of intersection of the cylinders $x^2+y^2=25$ and $y^2+z^2=20$ at the point $(3,4,2)$.

This was their solution:

enter image description here

I'm confused about the first part where they mention that the projection is contained in the circle $x^2+y^2=25$. I graphed the two cylinders using GeoGabra and got:

enter image description here

From the looks of it, the projection of the intersection doesn't look like a circle. Unless they literally meant contained within that circle which I guess is true but could we have also said that this curve was contained within $x^2+y^2=36$ and working with $x=6$ $\cos{t}$ and $y=6$ $\sin{t}$ for the rest of the problem and have that work as well?

I'm having a hard time wrapping my head around using the projection when the entire trace isn't covered in the intersection like in the question above where there is a small portion in red that isn't intersected by the other cylinder. I hope my question makes sense, please let me know if there are parts I should expand on.


I think you are correct in saying that they literally meant that the curve is "contained" in the circle $x^2+y^2=25$. More precisely, $C$ lies on the boundary of the cylinder with radius $5$. Note that each cross-section of the cylinder is the circle $x^2+y^2=25$.

  • $\begingroup$ But it would also be contained within the circle $x^2+y^2=36$ right? In that case, could I also have done the entire question where instead of having $5cos(t)$ and $5sin(t)$, have it as $6cos(t)$ and $6sin(t)$ where it'll eventually lead to an equivalent answer? $\endgroup$ – BoilingKettle Dec 26 '18 at 6:27
  • $\begingroup$ I'm not too sure where you're getting $x^2+y^2=36$. To solve the problem, you need the curve of intersection of two cylinders. More specifically, we want the curve of intersection on the boundary of these cylinders. If $x^2+y^2 =36$, we would be looking at a cylinder of radius $6$, whereas the original problem considers a cylinder of radius $5$. $\endgroup$ – J. Pistachio Dec 26 '18 at 6:32
  • $\begingroup$ Ah, my bad, when I was thinking of contained, i was thinking of a disc of radius 5 rather than circle of radius 5. I think I understand now, thanks. $\endgroup$ – BoilingKettle Dec 26 '18 at 6:36

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