These points would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation.įinding Zeros and Maximum Values for a Polar Equation We may find additional information by calculating values of r r when θ = 0. Similarly, the maximum value of the sine function is 1 when θ = π 2, θ = π 2, and if our polar equation is r = 5 sin θ, r = 5 sin θ, the value θ = π 2 θ = π 2 will yield the maximum | r |. The maximum value of the cosine function is 1 when θ = 0, θ = 0, so our polar equation is 5 cos θ, 5 cos θ, and the value θ = 0 θ = 0 will yield the maximum | r |. Consider r = 5 cos θ r = 5 cos θ the maximum distance between the curve and the pole is 5 units. įor many of the forms we will encounter, the maximum value of a polar equation is found by substituting those values of θ θ into the equation that result in the maximum value of the trigonometric functions. We use the same process for polar equations. Recall that, to find the zeros of polynomial functions, we set the equation equal to zero and then solve for x. To find the zeros of a polar equation, we solve for the values of θ θ that result in r = 0. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect. In these instances, we can confirm that symmetry exists by plotting reflecting points across the apparent axis of symmetry or the pole. However, failing the symmetry tests does not necessarily indicate that a graph will not be symmetric about the line θ = π 2, θ = π 2, the polar axis, or the pole. Passing one or more of the symmetry tests verifies that symmetry will be exhibited in a graph. The equation has failed the symmetry test, but that does not mean that it is not symmetric with respect to the pole. For example, suppose we are given the equation r = 2 sin ( 3 θ ). We replace ( r, θ ) ( r, θ ) with ( − r, θ ) ( − r, θ ) to determine if the tested equation is equivalent to the original equation. In the third test, we consider symmetry with respect to the pole (origin). The graph of this equation exhibits symmetry with respect to the polar axis. r = 1 − 2 cos θ Even/Odd identity r = 1 − 2 cos θ r = 1 − 2 cos ( − θ ) Replace ( r, θ ) with ( r, − θ ). For example, suppose we are given the equation r = 1 − 2 cos θ. We replace ( r, θ ) ( r, θ ) with ( r, − θ ) ( r, − θ ) or ( − r, π − θ ) ( − r, π − θ ) to determine equivalency between the tested equation and the original. In the second test, we consider symmetry with respect to the polar axis ( x x-axis). This equation exhibits symmetry with respect to the line θ = π 2. − r = −2 sin θ Identity: sin ( − θ ) = − sin θ. r = 2 sin θ − r = 2 sin ( − θ ) Replace ( r, θ ) with ( − r, − θ ). R = 2 sin θ − r = 2 sin ( − θ ) Replace ( r, θ ) with ( − r, − θ ). For example, suppose we are given the equation r = 2 sin θ r = 2 sin θ We replace ( r, θ ) ( r, θ ) with ( − r, − θ ) ( − r, − θ ) to determine if the new equation is equivalent to the original equation. In the first test, we consider symmetry with respect to the line θ = π 2 θ = π 2 ( y-axis). Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of r ) r ) to determine the graph of a polar equation. By performing three tests, we will see how to apply the properties of symmetry to polar equations. If an equation has a graph that is symmetric with respect to an axis, it means that if we folded the graph in half over that axis, the portion of the graph on one side would coincide with the portion on the other side. Symmetry is a property that helps us recognize and plot the graph of any equation. All points that satisfy the polar equation are on the graph. Recall that the coordinate pair ( r, θ ) ( r, θ ) indicates that we move counterclockwise from the polar axis (positive x-axis) by an angle of θ, θ, and extend a ray from the pole (origin) r r units in the direction of θ. Just as a rectangular equation such as y = x 2 y = x 2 describes the relationship between x x and y y on a Cartesian grid, a polar equation describes a relationship between r r and θ θ on a polar grid.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |