
How do you use $\cos \theta $ and $\sin \theta $ with positive coefficients to parameterize the intersection of the surfaces ${x^2} + {y^2} = 25$ and $z = 3{x^2}$ ?
Answer
534.3k+ views
Hint: The given question requires us to find the intersection point of the two surfaces and represent that point of intersection in terms of a parameter. The point of intersection of the curves is the point that satisfies both the curves ${x^2} + {y^2} = 25$ and $z = 3{x^2}$ simultaneously. So, we have to find such a point that satisfies all the conditions and can be represented in parametric form.
Complete step by step answer:
Let us consider the equation of a circle in the standard form whose centre is at $\left( {h,k} \right)$ and radius is r units.
So, we have,${(x - h)^2} + {(y - k)^2} = {r^2} - - - - (1)$
Now, we are given an equation of the circle as ${x^2} + {y^2} = 25$. So, on comparing the two equations, we can deduce that the center of the circle is $\left( {0,0} \right)$ and radius of the circle is $5$ units.
Now, we know that the parametric form of the points lying on a particular circle is $\left( {r\cos \theta ,r\sin \theta } \right)$.
So, the parametric form of the points lying on the circle ${x^2} + {y^2} = 25$ is of the form $\left( {5\cos \theta ,5\sin \theta } \right)$.
Now, we have got the parametric coordinates of any point lying on the circle whose equation is given to us.
Now, we have to consider the second equation given to us.
$z = 3{x^2}$
For finding the intersection point of the two surfaces, we have to solve the two equations simultaneously such that the solution point should satisfy both the equations. So, we have $x = 5\cos \theta $ and $y = 5\sin \theta $.
Now, $z = 3{x^2}$
$ \Rightarrow z = 3{\left( {5\cos \theta } \right)^2}$
$ \Rightarrow z = 75{\cos ^2}\theta $
So, the intersection point of the two surfaces is of the form $\left( {5\cos \theta ,5\sin \theta ,75{{\cos }^2}\theta } \right)$.
Note: The parametric coordinates of any point lying on the circle can be found easily. Intersection point is the point between two curves or surfaces that is common to both of them and satisfies both the equations.
Complete step by step answer:
Let us consider the equation of a circle in the standard form whose centre is at $\left( {h,k} \right)$ and radius is r units.
So, we have,${(x - h)^2} + {(y - k)^2} = {r^2} - - - - (1)$
Now, we are given an equation of the circle as ${x^2} + {y^2} = 25$. So, on comparing the two equations, we can deduce that the center of the circle is $\left( {0,0} \right)$ and radius of the circle is $5$ units.
Now, we know that the parametric form of the points lying on a particular circle is $\left( {r\cos \theta ,r\sin \theta } \right)$.
So, the parametric form of the points lying on the circle ${x^2} + {y^2} = 25$ is of the form $\left( {5\cos \theta ,5\sin \theta } \right)$.
Now, we have got the parametric coordinates of any point lying on the circle whose equation is given to us.
Now, we have to consider the second equation given to us.
$z = 3{x^2}$
For finding the intersection point of the two surfaces, we have to solve the two equations simultaneously such that the solution point should satisfy both the equations. So, we have $x = 5\cos \theta $ and $y = 5\sin \theta $.
Now, $z = 3{x^2}$
$ \Rightarrow z = 3{\left( {5\cos \theta } \right)^2}$
$ \Rightarrow z = 75{\cos ^2}\theta $
So, the intersection point of the two surfaces is of the form $\left( {5\cos \theta ,5\sin \theta ,75{{\cos }^2}\theta } \right)$.
Note: The parametric coordinates of any point lying on the circle can be found easily. Intersection point is the point between two curves or surfaces that is common to both of them and satisfies both the equations.
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