Category: Differential Calculus, Analytic Geometry, Algebra, Trigonometry
"Published in Newark, California, USA"
Find the angle of intersection between the two curves:
Solution:
Consider the given pair of two curves above
The first thing that we need to do is to get their point of intersection by solving the systems of equation as follows
but
Hence, the above equation becomes
The value of y is
The point of intersection of the given two curves is P(1, 2).
Consider the first given equation of a curve
Take the derivative on both sides of the equation with respect to x by implicit differentiation, we have
The
slope of a curve is equal to the first derivative of the equation of a
curve with respect to x. In this case, dy/dx is the slope of a curve. Actually, the first curve is a straight line.
Since the right side of the equation contains coefficient only, the slope of the first curve is
Consider the second given equation of a curve
Take the derivative on both sides of the equation with respect to x by implicit differentiation, we have
The
slope of a curve is equal to the first derivative of the equation of a
curve with respect to x. In this case, dy/dx is the slope of a curve.
To get the value of the slope of a curve at their point of intersection, substitute x = 1 and y = 2 at the equation above, we have
The slope of a curve at their point of intersection is equal to the slope of tangent line that passes thru also at their point of intersection. Hence,
Therefore, the angle between two curves at their point of intersection is
or
Category: Differential Calculus, Analytic Geometry, Algebra
"Published in Newark, California, USA"
Find the equations of the tangent line and the normal line that passes through the point P(-3, -1) for the curve:
Solution:
The
first thing that we have to do is to check if the given point is
included in the curve. Consider the given equation of a curve
The given equation represents a conic section because both x and y are second degree but we cannot identify the exactly type of a conic section because of the xy term. To find out what type of a conic section is, we can use the discriminant formula as follows
where B is the coefficient of xy, A is the coefficient of x² and C is the coefficient of y². Substitute the values of A, B, C to the above equation, we have
Since the value of a discriminant is greater than zero, then the given conic section is hyperbola in which its transverse and conjugate axes are not parallel to x and y axes.
From the coordinates of the given point, substitute x = -3 and y = -1 at the given equation, we have
Since both sides of the equation are equal, then the given point is included in the curve.
Again, consider the given equation of a curve
Take the derivative on both sides of the equation with respect to x by implicit differentiation, we have
The
slope of a curve is equal to the first derivative of the equation of a
curve with respect to x. In this case, dy/dx is the slope of a curve.
To get the value of the slope of a curve at the given point, substitute x = 3 and y = 1 at the equation above, we have
The slope of a curve at the given point is equal to the slope of tangent line that passes thru also at the given point. Hence,
Therefore, the equation of a tangent line is
Normal line is also a straight line which is perpendicular to tangent line at the point of tangency. Hence,
Therefore, the equation of a normal line is
