I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).
The arc length can be computed with a line integral, but first we'll need a parameterization for
. This is easy enough to do. First fix any one variable. For convenience, choose
.
Now,
, and
. The intersection is thus parameterized by the vector-valued function
where
. The arc length is computed with the integral
Some rewriting:
Complete the square to get
So in the integral, you can substitute
to get
Next substitute
, so that the integral becomes
This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):
So the arc length is
Answer:
x = 31.8
Step-by-step explanation:
From the picture attached,
By applying Pythagoras theorem in right ΔADC,
AC² = AD² + CD²
(22)² = AD² + (16)²
AD² = 484 - 256
AD = √228
= 2√57
Now we apply Pythagoras theorem in right ΔADB,
AB² = AD² + DB²
x² = 228 + (44 - 16)²
x² = 228 + 784
x² = 1012
x = √1012
= 31.81
≈ 31.8
108 is the answer to your question
Answer:
28
Step-by-step explanation:
You must multiply first using BODMAS oe BIDMAS, then add on.
18x2= 36
36-8= 28