To analyze: The graph given by the polar equation r=−3cos4θ .
The equation represents a rose curve with 8 petals. Domain of the graph is (−∞,∞) .
The range is [−3,3] . The graph is continuous and symmetric about both the axes and the origin. Maximum value of r is 3 . There are no asymptotes to the graph.
Given information:
The given polar equation is r=−3cos4θ .
Calculation:
Plot the graph for the polar equation.
Figure (1)
From the figure (1), it can be observed that the equation represents a rose curve with 8 petals.
The equation is defined for all real numbers of θ in the domain (−∞,∞) .
Range of cos4θ is, −1≤cos4θ≤1 .
The range of r=2sin3θ will be [−3,3] considering the 1 as maximum value and −1 as minimum value for cos4θ .
The graph is continuous. The rose curve is symmetric about both the axes and the origin.
Since r values are bounded between [−3,3] , it is a bounded graph.
The maximum value of r is |−3|=|3|=3 .
There are no asymptotes to the graph.
Therefore, the polar equation r=−3cos4θ is analyzed with the help of figure (1).
Chapter 6 Solutions
PRECALCULUS:GRAPHICAL,...-NASTA ED.
- 3. Solve the inequality, and give your answer in interval notation. - (x − 4)³ (x + 1) ≥ 0arrow_forward1. Find the formula to the polynomial at right. Show all your work. (4 points) 1- 2 3 сл 5 6 -4 -3 -2 -1 0 2 3arrow_forward2. Find the leading term (2 points): f(x) = −3x(2x − 1)²(x+3)³ -arrow_forward
- 1- √ √ √³ e³/√xdy dx 1 cy² 2- √ √² 3 y³ exy dx dy So 3- √ √sinx y dy dx 4- Jo √² Sy² dx dyarrow_forwardA building that is 205 feet tall casts a shadow of various lengths æ as the day goes by. An angle of elevation is formed by lines from the top and bottom of the building to the tip of the shadow, as de seen in the following figure. Find the rate of change of the angle of elevation when x 278 feet. dx Round to 3 decimal places. Γ X radians per footarrow_forwardUse the information in the following table to find h' (a) at the given value for a. x|f(x) g(x) f'(x) g(x) 0 0 0 4 3 1 4 4 3 0 2 7 1 2 7 3 3 1 2 9 4 0 4 5 7 h(x) = f(g(x)); a = 0 h' (0) =arrow_forward
- Use the information in the following table to find h' (a) at the given value for a. x f(x) g(x) f'(x) g'(x) 0 0 3 2 1 1 0 0 2 0 2 43 22 4 3 3 2 3 1 1 4 1 2 0 4 2 h(x) = (1/(2) ²; 9(x) h' (3)= = ; a=3arrow_forwardThe position of a moving hockey puck after t seconds is s(t) = tan a. Find the velocity of the hockey puck at any time t. v(t) ===== b. Find the acceleration of the puck at any time t. -1 a (t) = (t) where s is in meters. c. Evaluate v(t) and a (t) for t = 1, 4, and 5 seconds. Round to 4 decimal places, if necessary. v (1) v (4) v (5) a (1) = = = = a (4) = a (5) = d. What conclusion can be drawn from the results in the previous part? ○ The hockey puck is decelerating/slowing down at 1, 4, and 5 seconds ○ The hockey puck has a constant velocity/speed at 1, 4, and 5 seconds ○ The hockey puck is accelerating/speeding up at 1, 4, and 5 secondsarrow_forwardquestion 8arrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning





