R = sec(θ) − 2cos(θ), where -π/2 < θ < π/2
1 answer:
Answer:
y = (x/(1-x))√(1-x²)
Step-by-step explanation:
The equation can be translated to rectangular coordinates by using the relationships between polar and rectangular coordinates:
x = r·cos(θ)
y = r·sin(θ)
x² +y² = r²
__
r = sec(θ) -2cos(θ)
r·cos(θ) = 1 -2cos(θ)² . . . . . . . . multiply by cos(θ)
r²·r·cos(θ) = r² -2r²·cos(θ)² . . . multiply by r²
(x² +y²)x = x² +y² -2x² . . . . . . . substitute rectangular relations
x²(x +1) = y²(1 -x) . . . . . . . . . . . subtract xy²-x², factor
y² = x²(1 +x)/(1 -x) = x²(1 -x²)/(1 -x)² . . . . multiply by (1-x)/(1-x)
__
The attached graph shows the equivalence of the polar and rectangular forms.
You might be interested in
Applying our power rule gets us our first derivative,
simplifying a little bit,
looking for critical points,
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
and take our second derivative.
Looking for inflection points,
Again, pulling out the smaller power of x, and fractional part,
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
simplifying a little bit,
looking for critical points,
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
and take our second derivative.
Looking for inflection points,
Again, pulling out the smaller power of x, and fractional part,
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.