Answer
\[
x(t)=3-4t, \quad y(t)=4+3t, \quad z(t)=2-6t
\]
Work Step by Step
We have the cylinders:
\[
x^2+y^2=25 \quad and \quad y^2+z^2=20
\]
Reparametrizing the first cylinder \(x^2+y^2=25\) we have:
\[
x=5\cos t \\
y=5\sin t
\]
Isolating \(z\) at the equation of cylinder \( y^2+z^2=20\):
\[
z=\sqrt{20-y^2}
\]
Substituting \(y\):
\[
z=\sqrt{20-(5\sin t)^2}=\sqrt{20-25\sin^2t}
\]
At \((3,4,2)\) we have:
\[
5\cos t=3 \implies \cos t=\frac{3}{5}, \quad 5\sin t = 4 \implies \sin t=\frac{4}{5}
\]
Now that we have all the factors written as dependent variables, we can derivate the vector corresponding to the curve.
\[
\vec{r}(t)=\langle 5 \cos t, 5 \sin t, \sqrt{20-25\sin^2t} \rangle
\]
Derivating, we have:
\[
\vec{r}'(t)=\langle -5 \sin t, 5 \cos t, \frac{-25 \sin t\cos t}{\sqrt{20-25\sin^2t}}\rangle
\]
Substituting \(\cos t = \frac{3}{5}\) and \(\sin t = \frac{4}{5}\) we get:
\[
\frac{dx}{dt} = -5 \cdot \frac{4}{5} = -4 \\
\frac{dy}{dt} = 5 \cdot \frac{3}{5} = 3 \\
\frac{dz}{dt} = \frac{-25 \cdot \frac{4}{5} \cdot \frac{3}{5}}{\sqrt{20-25\cdot (\frac{4}{5})^2}} = \frac{-25 \cdot \frac{12}{25}}{\sqrt{20-25\cdot \frac{16}{25}}} =\frac{-12}{\sqrt{20-16}} = \frac{-12}{2} =-6
\]
Thus the tangent vector is:
\[
\vec{T}=(-4,3,-6)
\]
The vector equation for the tangent line at \(P(3,4,2)\) is:
\[
\vec{r}'(t) = \vec{r_0}+t\vec{T}
\]
Where \(\vec{r_0}=(3,4,2)\) and \(\vec{T}=(-4,3,-6)\). So the parametric equations of the tangent line are:
\[
x(t)=3-4t, \quad y(t)=4+3t, \quad z(t)=2-6t
\]
