Polar etc

Parametric equation

Equation where the x and y coordinate are expressed separately in terms of a third variable usually t.

vector valued function

In practice there is little difference between a vector valued function and a parametric function. Except that a vector valued function gives a vector which points to a position usually in 2d or 3d space

Polar Coordinates

Polar coordinates are like normal coordinates in that they have a point but different in the sense that they are measured in distance from the origin and angle. This means graphs of polar functions have a rather unique geometry. A point in polar coordinates is written as: (r, θ) r is the distance from the origin (also called the pole). θ (theta) is the angle, measured in degrees or radians, from the positive x-axis (also called the polar axis). (when putting into knowt put picw/this.

How does one find dy/dx for a parametric function?

To find dy/dx or the rate of change of y with respect to the rate of change of x. We can interpret this as finding the derivative of our x value and putting the derivative of our y value over it. And that's exactly what we'll do.

How does one find the second dy/dx for a parametric function?

First find dy/dx then find the derivative of that. After put that derivative over the first derivative of x or dx.

How does one find the arc length of a parametric curve?

First find the derivative of x then find the derivative of y. After that just square them and add them together under the same square root. Finally integrate that from the 2 endpoints.

How does one differentiate a vector valued function?

vector valued functions are easy when it comes to differentiating for these just find the derivative of each coordinate function and write it in its place.The reason vector valued functions are differentiated like this is because they don't have a single output in the same way parametric ones do.

How does one find the magnitude of a particle's velocity vector at a certain time?

To do this one must first find the particle's velocity vector. When given such a question typically you'll be given an equation with it and told whether that function is of: position time,velocity or acceleration keep in mind: The derivative of position-time is velocity and the derivative of velocity is acceleration. Once you get the velocity vector or velocity function plug in the time value (t) for the respective coordinates. Square them all add them to each other then square root the result.

How does one find the rate of change of a particles y coordinate when it's moving along a curve at a point; when given the rate of change of x.

For these questions you'll usually get an equation like y= something so first differentiate that then multiply your result by the rate of change of x or dx (which is usually a constant.) From there just sub in the given x value at the given point and that'll usually tell you y= some constant. Finally all you have to do is sub dx and the y coordinate you found into the magnitude equation from earlier and that's your answer.

How does one find the rate of change of a particles y coordinate when it's moving along a curve at a point; when given the rate of change of x but not given a normal equation for y?

For these questions you'll typically get some equation putting x and y together like xy=some constant so all you'll have to do is implicitly differentiate making sure to put x' or y' whenever differentiating the respective value eventually once you've differentiated and subbed in the given x' value you'll get y' usually in the form of y'=some x/ some y or whatever. Just sub into that the given x and y coordinates and you should get a constant which is your answer.

How does one the magnitude of displacement between two times when given a velocity vector?

Since displacement is the area under a velocity curve we'll first find the integral of the given velocity vector which if you remember is easy as we just take the integrals of the functions and put them into their respective coordinate. After finding the integral set the bounds as the given values then we just need to do the definite integral of x and y then put that into the magnitude formula.

How does one find the magnitude of a particle's velocity vector as it moves along a curve at a certain point? When the curve isn't defined explicity.

Typically you'll be given a function for a curve and the rate at which the particle moves about the x axis. Together this is just a position time graph but we want the velocity vector so we'll have to differentiate. If the curve isn't defined explicitly (as y=something) you'll be given dx and will have to use that information to implicitly differentiate the curve. Once you've done that plug the given coordinates into what you have for y'. Finally plug x' and y' into the magnitude formula and that's your answer.

How do you differentiate x or y when given a polar function?

Because cos represents x in the unit circle and sin represents y to find the x or y coordinate at a certain point in a polar function you do sinthe function to find y and costhe function for x. Thus when you have to differentiate a polar function for x or y just take the function multiply it by either sin or cos then differentiate that. Keep in mind that because of the nature of polar functions theta and x are interchangable (although you shouldn't swap theta w/x).

How does one find the rate of change of an x or y coordinate when given a point?

First you can just differentiate the function for x or y then you must find theta. Remember that for polar coordinates there's no x value only theta which corresponds to an angle, so theta is whatever angle will align us with the given point. (The angles value is based off the unit circle definition. ex if given the point 0,2 theta is pi/2.)

How does one find the tangent line to a polar function at a certain theta value?

First one must find the coordinate of the point. To do this we multiply the function by sin to find the y value then cos to find the x value. After this we need to find the unified slope at the point. This is done by finding dy and plugging in theta then finding dx and plugging in theta. Put the numbers you get over each other and that's dy/dx or your unified slope. After that keep in mind slope intercept form; simply plug the y value, slope, and x value into their respective slots of the slope intercept form and that's your tangent line.

How does one find where the slope of a polar function has a vertical tangent line?

(For this to happen there must be an unbounded portion) Just set dx as equal to 0 write the reasonable answers. Then find dy check all those reasonable answers any answer which gives a non 0 value for dy and a 0 value for dx is the right answer.

How does one find the area enclosed by a polar function between two theta values?

Just use the formula shown.

How does one find the area enclosed by a polar function when unsure about the bounds of integration?

We can plug a small radian value like pi/4 into the equation and if it's on the positive side 0 is in the beginning of the positive side if pi/4 is negative the implication is self explanatory. From there we'll know what's what.

How does one find the area enclosed by two polar functions?

The key to these kinds of questions is just to imagine where the angles land and the best way to find the bounded area.

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when finding dy/dx for a value of a polar func just find y’ of x first then find x’ or x second