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area with parametric equationsin this section we will find out a formula for ascertaining the area under a parametric curve specified by the
vertical tangent for parametric equationsvertical tangents will take place where the derivative is not defined and thus well get vertical tangents at
horizontal tangents for parametric equations horizontal tangents will take place where the derivative is zero and meaning of this is that well get
find out the tangent lines to the parametric curve specified byx t5 - 4t3y t2at 04solution note that there is actually the potential for more than
derivative for parametric equationsdxdy dxdt dydt given dydt ne 0why would we wish to do this well remind that in
eliminate the parameter from the subsequent set of parametric equationsx t2 ty 2t - 1solutionone of the very easy ways to eliminate the parameter
a pair of straight lines are drawn through the origin forms with the line 2x3y6 an isoceles triangle right angled at origin find the equation of pair
parametric equations and curvestill to this point we have looked almost completely at functions in the form y f x or x h y and approximately all of
parametric equations and polar coordinatesin this part we come across at parametric equations and polar coordinates when the two subjects dont come
probability - applications of integralsin this final application of integrals that well be looking at we are going to look at probability
hydrostatic pressure and force - applications of integralsin this part we are going to submerge a vertical plate in water and we wish to know the
formulas of surface area - applications of integralss int 2piyds rotation about x-axiss int
arc length formulal int dswhereds radic 1 dydx2 dx
applications of integralsin this part were going to come across at some of the applications of integration it should be noted also that these
example for comparison test for improper integrals example find out if the following integral is convergent or divergentintinfin2 cos2 x x2
comparison test for improper integrals here now that weve seen how to actually calculate improper integrals we should to address one more topic about
following is some more common functions that are nice enough polynomials are nice enough for all xs if f x p x q x then fx will be nice enough
one-sided limits we do this along with one-sided limits as the name implies with one-sided limits we will just looking at one side of the point
velocity problem lets look briefly at the velocity problem several calculus books will treat it as its own problem in this problem we
rates of change or instantaneous rate of change now we need to look at is the rate of change problem it will turn out to be one of the most
rates of change and tangent lines in this section we will study two fairly important problems in the study of calculus there are two cause for
solve 2 ln radicx - ln 1 - x 2 solution firstly get the two logarithms combined in a single logarithm2 ln radicx - ln x - l 2ln radicx2 ln 1
solve 3 2 ln x 73 -4 solutionthis initial step in this problem is to get the logarithm by itself on one side of the equation along with a
solve following 4e13 x - 9e5-2 x 0 solutionhere the first step is to get one exponential on every side amp then well divide both sides by one of
solve following x - x e 5 x 2 0 solution the primary step is to factor an x out of both termsdo not divide an x from both termsnote as