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cardioids and limaconsthese can be split up into the following three cases1 cardioids r a a cos theta and r a a sin thetathese encompass a graph
determine equation of the tangent line to f x 4x - 8 radicx at x 16 solution we already know that the equation of a tangent line is specified
circles - common polar coordinate graphs let us come across at the equations of circles in polar coordinates1 r a this equation is saying that there
some interpretations of the derivativeexample is f x 2 x3 300 4 increasing decreasing or not changing at x -2 solution we already
differentiate following functions h t 2t5 t2- 5 t2we can
differentiate following functionsa f x 15x100 - 3x12 5x - 46b h x x pi - x radic2 solutiona f x 15x100 - 3x12 5x - 46in
lines- common polar coordinate graphsa few lines have quite simple equations in polar coordinates1 theta betawe are able to see that this is a
basic computation formulas next lets take a quick look at some basic computation formulas that will let us to actually compute some
differentiation formulas we will begin this section with some basic properties and formulas we will give the properties amp formulas in this
velocity recall that it can be thought of as special case of the rate of change interpretation if the situation of an object is specified by ft
polar to cartesian conversion formulasx r cos thetay r sin thetaconverting from cartesian is more or less easy lets first notice the
interpretations of the derivative before moving on to the section where we study how to calculate derivatives by ignoring the limits we were
surface area with parametric equationsin this final section of looking at calculus applications with parametric equations we will take a look at
arc length for parametric equationsl intbetaalpha radic dxdt2 dydt2 dtnote that we could have utilized the second formula for ds above is we had
find out the area under the parametric curve given by the following parametric equations x 6 theta - sin thetay 6 1 - cos theta0 le theta le
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