ASAP Refer to BC CALC Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions:

Question

Refer to BC CALC Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions:

GIVEN

the polar curve r is given by r(theta) = (theta)^2+7cos(theta), where 0 is less than or equal to theta which is less than or equal to pi.

QUESTIONS:

a) find the area above thex-axiss enclosed by x-axis and graph of r 

  b) for 0 is less than or equal to theta which is less than or equal to pi, there is one point P on the polar curve with x-coordinate 1. find the angle theta that corresponds to point P as well as the y-coordinate of P. Show the work that leads to your answer.

 c) A particle is moving along polar curve r such that its position is (x(t), y(t)) and (d *theta)/dt=1/2. Find dy/dt at the instant that theta= pi/6 and interpret the meaning of your answer in the context of this problem

(YOU DON'T NEED THE GRAPH)

Answer 1

Given:

r( θ ) = θ² + 7 cos( θ ) θ: [0, π] Or 0 ≤ θ ≤ π

Find:

a) A = ? un²

The area under the polar curve, bounded by the x-axis, and above the x-axis

b) θ = ?

When x = 0

c) dθ/dt = 1/2, θ = π/6, dy/dt = ? and interpret

Solution:

a)

When it says "bounded the x-axis and above x-axis", it means any crosses with the x-axis.

Without knowledge of fancy math lingo, the x-axis is "defined" by y = 0

The relation between polar and cartesian coordinates for the y-coordinate is:

y = r sin( θ ) So find the angle θ that causes y to go to zero

0 = ( θ² + 7 cos( θ ) ) sin( θ )

Warning: DO NOT DISTRIBUTE

The above is similar to finding roots of a polynomial (Ex: x² - 2x + 4 = ( x - 2 ) ( x - 2 ) = 0, x = 2)

With the "hard" part of separating them done in the "setup"

So, either:

θ² + 7 cos( θ ) = 0 OR sin( θ ) = 0

If you have a fancy enough calculator that can solve complex equations like the first one, you'll end up getting a no solution, meaning θ² + 7 cos( θ ) never equals zero (the polar curve never touches the origin because a radius of zero means it would lie on the origin, whatever angle(s) would have caused it to zero out). Thus, the simpler equation is where you determine your angles for zero-ing out the y-coordinate

sin( θ ) only equals zero (on these problems bounds of 0 and π) at (crazy enough), θ = 0 and θ = π

To find the area, evaluate:

A = ∫₀^π r( θ ) dθ = ∫₀^π ( θ² + 7 cos( θ ) ) dθ = θ³ / 3 + 7 sin( θ ) |₀^π = 1/3 ( π³ - 0³ ) + 7 * 0

a) ∴ A = π³ / 3 ≈ 10.34 un²

b)

x = r cos( θ ) 0 = ( θ² + 7 cos( θ ) ) cos( θ ) cos( θ ) = 0 ∴ θ = π / 2

c)

y = r sin( θ ) = ( θ² + 7 cos( θ ) ) sin( θ )

Using Chain Rule, Product Rule AND Power Rule:

dy/dt = [ ( θ² + 7 cos( θ ) ) * cos( θ ) + ( 2θ - 7 sin( θ ) ) sin( θ ) ] * dθ/dt

dy/dt = [ ( ( π/6 )² + 7 cos( π/6 ) ) * cos( π/6 ) + ( 2 * π/6 - 7 sin( π/6 ) ) sin( π/6 ) ] * 1/2

With a calculator on hand (making sure you watch those parantheses!!):

∴ dy/dt ≈ + 2.13 un / time unit

At the angle of π/6 radians, the y-coordinate is changing 2.13 units every unit of time (typically seconds but not restricted to seconds)

I hope this helps! Message me in the comments if you have any questions, comments, or concerns on how I handled the above!