MATH 2001September 9, 2013Assignment #1Due: September 251. Find the Cartesian equation of the curve defined by parametric equa-tionsx=2cost What curve is it?2. Consider the curvex=t4 Document Preview:...

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MATH 2001September 9, 2013Assignment #1Due: September 251. Find the Cartesian equation of the curve defined by parametric equa-tionsx=2cost
What curve is it?2. Consider the curvex=t4


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MATH 2001 September 9, 2013 Assignment #1 Due: September 25 1. Find the Cartesian equation of the curve de?ned by parametric equa- tions x=2cost cos2t, y=2sint sin2t. What curve is it? 2. Consider the curve 4 4 x = t 2t, y = t+t . (a) Graph the curve using a computational software package. (b) Use your graph to estimate the coordinates of the lowest point and the leftmost point on the curve. (c) Find the exact coordinates of your points from (b). 3. The astronomer Giovanni Cassini (1625-1712) studied the family of curves with polar equations 4 2 2 4 4 r 2c r cos2? +c a =0 where a and c are positive real numbers. These curves are called the ovals of Cassini even though they are oval shaped only for certain values of a and c.(Cassinithoughtthatthesecurvesmightrepresent planetary orbits better than Kepler’s ellipses). (a) Investigate the variety of shapes that these curves may have by considering di?erent values for a and c.Generateyourplotsus- ing a computational software package such as Maple (hint: see for plotting an equation of this type in Maple). (b) Determine how a and c are related to each other when the curve splits into two parts. 14. (a) Show that the equation of the tangent line to the parabola 2 y =4px at the point (x ,y)canbewrittenas 0 0 y y=2p(x+x ) 0 0 (b) What is the x-intercept of this tangent line? Use this fact to draw the tangent line. 5. Find the volume of the solid generated by rotating the top half (y 0) of the ellipse 2 2 x y + =1 2 2 a b about the x-axis. 2

Answered Same DayDec 24, 2021

Answer To: MATH 2001September 9, 2013Assignment #1Due: September 251. Find the Cartesian equation of the curve...

David answered on Dec 24 2021
116 Votes
Sol: (1)
 
2 2
2
2
2 2 2 2
2 2
5
cos
4
Put the value of cos in , we get
2cos 2cos 1
5 5
2 2 1
4 4
5

2
x y
t
t x
x t t
x y x y
x
x y
x
 

  
       
           
  
  
 
  
2
2 2
2 2
4 4 2 2 2 2
2 2
4 4 2 2 2 2
2 2
1
5 1
8
5 1
25 10 2 10 1
2 8
5 1
25 10 2 10 8
2 8
20 4 4 25

x y
x y
x x y x x y y
x y
x x y x x y y
x y x
x
  
      
  
    
          
  
  
        
 
   

4 4 2 2 2 2
2 2 4 4 2 2
4 4 2 2 2 2
10 2 10 8
8
8 20 6 6 25 2 8
2 6 6 8 17 0
Name of the curve: Circle
y x x y y
x x y x y x y
x y x y x y x
    
       
      
Sol: (2) (a)
(b)
 
3
3
It looks like the leftmost point of the curve is 1.053,1.472 .
The leftmost part of the graph is where 0.
4 2
4 2 0

dx
dt
dx
t
dt
t


 
 
3
3
3
4...
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