Finding a vector equation for the tangent line of curve formed by the intersection of two cylinders

Question

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:

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:

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.

Answer 1

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 $x^2+y^2=25$.

Answer 2

I also was confused about the process of the book, but after investigating other sources, I found that the process is somewhat incomplete. To solve these problems, we have to do two steps:

1. Match the two equations.
2. Parameterize the intersection equation.

So, for the first step: We have:

x^2 +y^2 = 25 (1)

z^2 +y^2 = 20 (2)

You can make (1) match to (2) by subtracting five from both sides of (1)

x^2+y^2 - 5=20 (3)

Substituting 20 for (3) and then operate

x^2 + y^2 - 5= z^2 + y^2 (4)

x^2 -z^2 = 5 (5)

The second step is to parameterize. I chose x = t. Then

x = t

You can find z from (5)

z = sqrt(x^2 - 5), but x=t -> z = sqrt(t^2 - 5)

And find y from (1) or (2). I chose (1), then

y = sqrt(25 - x^2), but x = t -> y = sqrt(25 - t^2)

Finally, you have:

x =t

y = sqrt(25 - t^2)

z = sqrt(-5 + t^2)

This link https://www.kristakingmath.com/blog/vector-function-for-curve-of-intersection can help to solve this type of problem. And also remember to use Geogebra3D to get a better idea of what you are looking for.

PD: I apologize if sometimes I "speak weirdly"; English is not my mother tongue.

Answer 3

We have

$$ \cases{\mathcal{C}_1\to x^2+y^2=25 \\ \mathcal{C}_2\to y^2+z^2 = 20 \\ \mathcal{L}\to p = p_0 + \lambda \vec v \\ \|\vec v\|^2 = 1} $$

where $\mathcal{C}_1,\mathcal{C}_2$ are cylinders and $\mathcal{L}$ is the tangent line. Also

$$ \cases{ p = (x, y, z)\\ p_0 = (x_0,y_0,z_0) = (3,4,2)\\ \vec v = (v_x,v_y, v_z) } $$

After substitution of $\mathcal{L}$ on $\mathcal{C}_1,\mathcal{C}_2$

$$ \cases{(x_0 + \lambda v_x)^2+(y_0+\lambda v_y)^2 = 25\\ (y_0+\lambda v_y)^2+ (z_0+\lambda v_z)^2 = 20} $$

Solving for $\lambda$ we have

$$ (v_x^2+v_y^2)\lambda^2+2(x_0v_x+y_0v_y)\lambda + x_0^2+y_0^2=25)\Rightarrow \lambda = \frac{2(x_0v_x+y_0v_y)\pm\sqrt{4(x_0v_x+y_0v_y)^2-4(v_x^2+v_y^2)(x_0^2+y_0^2-25)}}{v_x^2+v_y^2} $$

but at tangency is required that

$$ 4(x_0v_x+y_0v_y)^2-4(v_x^2+v_y^2)(x_0^2+y_0^2-25)=0 $$

considering also the second equation we have finally

$$ \cases{ (x_0v_x+y_0v_y)^2-(v_x^2+v_y^2)(x_0^2+y_0^2-25)=0\\ (y_0v_y+z_0 v_z)^2-(v_y^2+v_z^2)(y_0^2+z_0^2-20) = 0 \\ v_x^2+v_y^2+v_z^2= 1 } $$

and with numeric values

$$ \cases{ 3v_x+4v_y = 0 \\ 4 v_y+2 v_z = 0 \\ v_x^2+v_y^2-v_z^2 = 1 } $$

hence

$$ \vec v =\frac{1}{\sqrt{61}} (4,-3,6) $$