consider the polar graph §æ =(2/3)¦Ð. (¦Ð=pie)
find the value of N such that r=N§æ, where 0¡Ü §æ ¡ÜN¦Ð has
a)no solution
b)one solution
c)two solution
i think i have figured the first question out.since it's no solution then N must be N<(2/3) because the equation is §æ =(2/3)¦Ð.
but i have no idea how to solve the following two questions!! can someone please help me with this ??
thanks!
To find the value of N such that the equation r = Nθ has no solution, we need to consider the range of values for θ such that 0 ≤ θ ≤ Nπ.
A) No solution:
For the equation to have no solution, there should be no values of θ that satisfy the range condition. This occurs when N < 2/3, as you correctly stated.
B) One solution:
For the equation to have one solution, there should be exactly one value of θ that satisfies the range condition. This occurs when N = 2/3.
C) Two solutions:
For the equation to have two solutions, there should be exactly two values of θ that satisfy the range condition. This occurs when N > 2/3.
So, the value of N such that the equation r = Nθ has no solution is N < 2/3, one solution is N = 2/3, and two solutions is N > 2/3.
To find the value of N such that the equation has a certain number of solutions, we can examine the relationship between N and the graph of §æ = (2/3)¦Ð.
First, let's plot the graph of §æ = (2/3)¦Ð.
To do this, we can create a table of values for §æ (theta) and find corresponding values for r (radius) using the equation r = N§æ.
Let's consider three values of §æ: §æ = 0, §æ = π/2, and §æ = π.
For §æ = 0, the equation becomes:
r = N * 0
r = 0
For §æ = π/2, the equation becomes:
r = N * (π/2)
r = (Nπ)/2
For §æ = π, the equation becomes:
r = N * π
r = Nπ
Now, let's examine the different cases:
a) No solution:
If there are no solutions, it means that the graph of §æ = (2/3)¦Ð does not intersect with any ray from the origin. Looking at the graph, we can see that this happens when N < 2/3.
b) One solution:
If there is one solution, it means that the graph of §æ = (2/3)¦Ð intersects with exactly one ray from the origin. Looking at the graph, we can see that this happens when N = 2/3.
c) Two solutions:
If there are two solutions, it means that the graph of §æ = (2/3)¦Ð intersects with two different rays from the origin. Looking at the graph, we can see that this happens when N > 2/3.
Therefore, the answer is:
a) No solution: N < 2/3
b) One solution: N = 2/3
c) Two solutions: N > 2/3