tangents with polar coordinateshere we now


Tangents with Polar Coordinates

Here we now require to discuss some calculus topics in terms of polar coordinates.

We will begin with finding tangent lines to polar curves.  In this case we are going to suppose that the equation is in the form r = f (θ). Along with the equation in this form we can in fact make use of the equation for the derivative dy/dx.  We derived while we looked at tangent lines along with parametric equations. Though, to do this requires us to come up with a set of parametric equations to present the curve. In fact this is pretty easy to do.

From our work in the preceding section we have the subsequent set of conversion equations for going from polar coordinates to Cartesian coordinates.

x = r cos θ

y = r sin θ

Now here, we'll use the fact that we were assuming that the equation is in the form r = f (θ).

Substituting this into these equations provides the following set of parametric equations (along with θ like the parameter) for the curve.

From our work in the preceding section we have the subsequent set of conversion equations for going from polar coordinates to Cartesian coordinates.

x= r cosθ

y = r sinθ

Here now, we'll make use of the reality that we're assuming that the equation is in the form r = f (θ).  Substituting this into these equations provides the subsequent set of parametric equations (with θ like the parameter) for the curve.

x = f (θ) cos θ

 y = f (θ) sin θ

 Now, we will require the following derivatives.

 dx / dθ = f' (θ) cosθ - f (θ) sin θ

= dr / dθ (cosθ) - rsinθ

dy/dθ = f′ (θ) sinθ + f (θ) cosθ

 = dr/dθ (sinθ) + r cosθ

Request for Solution File

Ask an Expert for Answer!!
Mathematics: tangents with polar coordinateshere we now
Reference No:- TGS0264527

Expected delivery within 24 Hours