Answer:
a) 25.15
b)
x = 1
y = t
z = (4pi)^2 + t *(8pi) = 4pi(4pi + 2t)
c) (x,y) = (1, -2pi)
Step-by-step explanation:
a)
First lets calculate the velocity, that is, the derivative of c(t) with respect to t:
v(t) = (-sin(t), cos(t), 2t)
The velocity at t0=4pi is:
v(4pi) = (0, 1, 8pi)
And the speed will be:
s(4pi) = √(0^2+1^2+ (8pi)^2) = 25.15
b)
The tangent line to c(t) at t0 = 4pi has the parametric form:
(x,y,z) = c(4pi) + t*v(4pi)
Since
c(4pi) = (1, 0, (4pi)^2)
The tangent curve has the following components:
x = 1
y = t
z = (4pi)^2 + t *(8pi) = 4pi(4pi + 2t)
c)
The intersection with the xy plane will occurr when z = 0
This happens at:
t1 = -2pi
Therefore, the intersection will occur at:
(x,y) = (1, -2pi)
Answer:
To me it may be a
hope this help can i get a branliest if you got it right and tell me if it is
(–23) – 45 – (–98) =–23 – 45 +98=-68+98=30
Answer:
<h3>(-4, 1)</h3>
Step-by-step explanation:
-x-3=y, therefore x = -y-3
-3x - 8y = 4
Find the value of y
Substitute the x in -3x - 8y = 4 with -y-3
We get
-3·(-y-3) - 8y = 4
3y + 9 - 8y = 4
-5y = 4-9
-5y = -5
<h3>y = 1</h3>
_______________
Find the value of x
x = -y-3
x = -1-3
<h3>x = -4</h3>
_______________
Answer (x, y) =
<h3>(-4, 1)</h3>
_____________________
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