C=πd where C=circumference, π=the constant pi, and d=diamter
d=C/π and since we are told that C=9π
d=9π/π
d=9
Answer:
Perpendicular
Skew
Parallel
Step-by-step explanation:
First one : the lines intersect and they form right angles
The second one : they don't intersect and appear on separate planes
The last : they are both sides of a rectangle on the same plane
Let
. The gradient of
at the point (1, 0, 0) is the normal vector to the surface, which is also orthogonal to the tangent plane at this point.
So the tangent plane has equation
![\nabla f(1,0,0)\cdot(x-1,y,z)=0](https://tex.z-dn.net/?f=%5Cnabla%20f%281%2C0%2C0%29%5Ccdot%28x-1%2Cy%2Cz%29%3D0)
Compute the gradient:
![\nabla f(x,y,z)=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)=\left(-\dfrac1{x+y},-\dfrac1{x+y},1\right)](https://tex.z-dn.net/?f=%5Cnabla%20f%28x%2Cy%2Cz%29%3D%5Cleft%28%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D%2C%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%2C%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20z%7D%5Cright%29%3D%5Cleft%28-%5Cdfrac1%7Bx%2By%7D%2C-%5Cdfrac1%7Bx%2By%7D%2C1%5Cright%29)
Evaluate the gradient at the given point:
![\nabla f(1,0,0)=(-1,-1,1)](https://tex.z-dn.net/?f=%5Cnabla%20f%281%2C0%2C0%29%3D%28-1%2C-1%2C1%29)
Then the equation of the tangent plane is
![(-1,-1,1)\cdot(x-1,y,z)=0\implies-(x-1)-y+z=0\implies\boxed{z=x+y-1}](https://tex.z-dn.net/?f=%28-1%2C-1%2C1%29%5Ccdot%28x-1%2Cy%2Cz%29%3D0%5Cimplies-%28x-1%29-y%2Bz%3D0%5Cimplies%5Cboxed%7Bz%3Dx%2By-1%7D)
2 is greater than one and 3 and 4 and 5