Category: Differential Calculus, Algebra
"Published in Vacaville, California, USA"
Using the fact that
show that
Solution:
Consider the given equation above
Take the derivative on both sides of the equation with respect to x by quotient formula, we have
Therefore,
This website will show the principles of solving Math problems in Arithmetic, Algebra, Plane Geometry, Solid Geometry, Analytic Geometry, Trigonometry, Differential Calculus, Integral Calculus, Statistics, Differential Equations, Physics, Mechanics, Strength of Materials, and Chemical Engineering Math that we are using anywhere in everyday life. This website is also about the derivation of common formulas and equations. (Founded on September 28, 2012 in Newark, California, USA)
Showing posts with label Differential Calculus. Show all posts
Showing posts with label Differential Calculus. Show all posts
Sunday, March 2, 2014
Saturday, March 1, 2014
Second Derivative Problems - Quotient Formula
Category: Differential Calculus, Algebra
"Published in Vacaville, California, USA"
Find a formula for
Solution:
The first thing that we have to do is to get the derivative of u/v with respect to x where u and v are functions of x. Apply the derivative by quotient formula, we have
Take the derivative again of the above equation with respect to x by product formula, we have
where:
"Published in Vacaville, California, USA"
Find a formula for
Solution:
The first thing that we have to do is to get the derivative of u/v with respect to x where u and v are functions of x. Apply the derivative by quotient formula, we have
Friday, February 28, 2014
Second Derivative Problems - Product Formula
Category: Differential Calculus, Algebra
"Published in Vacaville, California, USA"
Find a formula for
Solution:
The first thing that we have to do is to get the derivative of uv with respect to x where u and v are functions of x. Apply the derivative by product formula, we have
Take the derivative again of the above equation with respect to x by product formula, we have
where:
"Published in Vacaville, California, USA"
Find a formula for
Solution:
The first thing that we have to do is to get the derivative of uv with respect to x where u and v are functions of x. Apply the derivative by product formula, we have
Take the derivative again of the above equation with respect to x by product formula, we have
where:
Thursday, February 27, 2014
Angle Between Two Curves, 4
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. If you subtract the first equation from the second equation, the left side of the equation will be equal to zero. Hence, the resulting equation is
The value of y is
The point of intersection of the given two curves is P(1, ½).
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.
To get the value of the slope of a curve at their point of intersection, substitute x = 1 and y = ½ 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,
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 = ½ 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
"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. If you subtract the first equation from the second equation, the left side of the equation will be equal to zero. Hence, the resulting equation is
The value of y is
The point of intersection of the given two curves is P(1, ½).
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.
To get the value of the slope of a curve at their point of intersection, substitute x = 1 and y = ½ 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,
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 = ½ 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
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