Category: Differential Calculus, Analytic Geometry, Algebra
"Published in Vacaville, California, USA"
Find the equations of the tangent line and the normal line that passes through the point P(0, 4) 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 circle because x and y are both in second degree and also have x and y as well. From the coordinates of the given point, substitute x = 0 and y = 4 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 = 0 and y = 4 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
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(1, 2) 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
From the coordinates of the given point, substitute x = 1 and y = 2 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 = 1 and y = 2 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
