Practice Problems: Calculus in Polar Coordinates Answers. For these problems, I want to convert from polar form parametrized Cartesian form, then differentiate and take the ratio y over x to get the slope, and then evaluate both the slope and the curve s location at the given point so I can construct a tangent line. (a) I will use the simple parametrization θ = t so we can write the polar curve as r = t. Now I can convert to Cartesian: then differentiate each with respect to t, The slope of the tangent line is given by x = r cos θ = t cos t y = r sin θ = t sin t, x = cos t t sin t y = sin t + t cos t. y sin t + t cos t x = cos t t sin t I want the tangent line at θ = π, which means t = π, so the slope is m = sin π + π cos π cos π π sin π = π = π The Cartesian coordinates of the point the curve passes through at t = π is So the tangent line is given by (x, y) = (π cos π, π sin π) = ( π, ) y = π(x ( π)) y = πx + π (b) I will use θ = t as my parametrization so the polar curve is: θ = t, r = sin t + cos(t) The Cartesian form of the curve is then x = r cos θ = [sin t + cos(t)] cos t y = r sin θ = [sin t + cos(t)] sin t Differentiating the x and y equations, x = [sin t + cos(t)]( sin t) + [cos t sin(t)] cos t y = [sin t + cos(t)] cos t + [cos t sin(t)] sin t The slope of a tangent line is then given by m = y x = [sin t + cos(t)] cos t + [cos t sin(t)] sin t [sin t + cos(t)]( sin t) + [cos t sin(t)] cos t
We want to evaluate this when theta =, which means t =, so m = [sin + cos ] cos + [cos sin ] sin [sin + cos ]( sin ) + [cos sin ] cos = = The point the curve passes through when t = is (x, y) = ([sin + cos ] cos, [sin + cos ] sin ) = (, ) The line of slope through the point (, ) is given, by y = x. The arc length formula for a parametric Cartesian curve is b (dx ) ( ) dy + dt dt dt (a) Using the θ = t parametrization, and converting to Cartesian form, a x = r cos θ = t cos t y = r sin θ = t sin t Differentiating with respect to t, x = t cos t t sin t y = t sin t + t cos t Then applying the length formula, 3 Expanding the squared terms, 3 (t cos t t sin t) + (t sin t + t cos t) dt 4t cos t t 3 sin t cos t + t 4 sin t + 4t sin t + t 3 sin t cos t + t 4 cos t dt Canceling and using sin t + cos t =, this simplifies to 3 If we do the substitution u = t + 4, du = t dt, 3 4 [ u du = ] 3 u 3/ 3 4 3 4t + t 4 dt = t 4 + t dt = ( (3) 3/ 3 ) (4)3/ = ( 3 3/ 4 3/).957 3 3
(b) As θ runs from to π, the radius is positive. At θ = π the radius returns to and becomes negative and the curve will retrace itself. That means the whole curve is described by the interval (, π). Using the θ = t parametrization, and converting to Cartesian form, x = r cos θ = 4 sin t cos t y = r sin θ = 4 sin t Differentiating with respect to t, x = 4 sin t + 4 cos t y = 8 sin t cos t Then applying the length formula, ( 4 sin t + 4 cos t ) + (8 sin t cos t) dt Expanding the squared terms, Combining like terms, 6 sin 4 t 3 sin t cos t + 6 cos 4 t + 64 sin t cos t dt 6 sin 4 t + 3 sin t cos t + 6 cos 4 t dt Factoring out a 6 inside the radical, and bringing it outside as a 4: 4 sin 4 t + sin t cos t + cos 4 t dt I can write the quantity under the radical as a square: (sin 4 t + cos t ) dt = 4 dt = 4π If you graph this curve, you will see it is a circle of radius centered at the point (, ), and so this circumference makes sense.
3. The arc length formula for a polar curve is b a r + ( ) dr dθ dθ (a) The radius is positive everywhere except at θ = π, but then it becomes positive again for the rest of the angles, so there is no chance that the curve will retrace itself before θ has gone from to π. At π the curve begins does retrace itself because cosine has period π. So the interval over which the entire curve is traced out is (, π). We re given the formula for r, we just need to differentiate it with respect to θ and plug both values into our length formula: = Applying sin θ + cos θ =, dr r = + cos θ dθ = sin θ ( + cos θ) + ( sin θ) dθ Now recall the half-angle formula: cos x = + cos θ + cos θ + sin θ dθ + cos θ dθ = + cos θ dθ + cos x and so cos x = + cos x (the absolute values are important because the square root is always positive, but cosine is negative at some points) If we do the substitution u = θ, du = dθ, π + cos(u) π + cos(u) du = 4 du = 4 cos u du To evaluate, we need to split into the interval where cosine is positive and where it is negative: ( / 4 ) π ( [ ] π/ [ ] ) π cos u du + cos u du = 4 sin u sin u = 4 ( + ) = 8 π/ π/ (b) We re given the formula for r and the interval, we just need to differentiate with respect to θ and plug both values into our length formula: r = e θ dr dθ = eθ (e θ ) + (e θ ) dθ = e θ dθ = e θ dθ This integral can be evaluated directly... [e θ] π = ( e π e ) = (e π )
4. For reference, the polar area formula is b a r dθ (a) Let s graph the curve to get an idea of the area we want: We want to allow θ to range over the entire interval from to π to cover the whole area, so π ( + cos θ) dθ = (4 + 4 cos θ + cos θ) dθ We can break this into three integrals, the first two of which are easy, and the last of which requires the half-angle identity: Evaluating, 4 dθ + 4 cos θ dθ + + cos θ dθ 4 [ ] π [ ] π 4θ + 4 sin θ + [ θ + sin(θ) 4 ] π = (4π ) + ( ) + 4 [(π + ) ( + )] = 4π + π = 9π In an eyeball estimate of the graph, the shaded area looks about the same size as (or slightly larger than) a circle of radius, whose area would be 4π, so our answer seems to make sense.
(b) Let s graph the curve and see what one leaf looks like: To get one leaf, we want θ to range from one place where r = until the next point when r = (from the center, out to the end of the leaf, then back to the center. Sine does this from to π, so 3θ should range from to π. That means θ runs from to π/3, and so that s our domain of integration. /3 sin (3θ) dθ We deal with even powers of sine and cosine using the half-angle identity: Evaluating, /3 cos 6θ dθ = 4 /3 [ ] π/3 [ sin(6θ) ] π/3 θ 4 6 ( π ) = ( ) = π cos 6θ dθ If I were to guess at the radius of a circle with the same area as the leaf, I might guess.5, which would give an area of π/6, which is pretty close to the π/ answer, so it seems reasonable.
5. The area formula is the same as before. To find the area between two curves, we we re now taking the difference of the outer curve s area and the inner curve s area. b (a) Let s graph the area we re trying to determine: a ( r out rin ) dθ This is not really an area between curves, but it s an area enclosed by both curves. Look at it from the point of view of the origin: In the blue region, the curve r = is the outside curve, and in the red region, the other curve, r = 4 cos θ is the outside curve (remember that we add up areas of fans made of tiny wedges with their points at the origin, so we have to draw lines from there to see what curve those lines run into first). So we need to know the angle at which the two curves intersect - to get this we set the radius equations equal to one another and solve for θ: = 4 cos θ, = cos θ, θ = π 3
We also need to know the angle when the second curve gets to the origin. If looks like π, but let s make sure. r = 4 cos θ will be when cos θ =, which happens at π. Let s just find the area of the top half of the graph, and then double it, using the symmetry of the graph to reduce our work-load /3 dθ + / π/3 (4 cos θ) dθ = /3 dθ + 8 / π/3 cos θ dθ For the second integral, we use the half-angle formula: π/3 / + cos(θ) π/3 dθ + 8 dθ = dθ + 4 π/3 / π/3 + cos(θ) dθ Evaluating, [ ] π/3 θ = [ + 4 ( π 3 ) + 4 ] π/ θ + sin(θ) π/3 ( π + sin π π 3 sin(π/3) ) Doubling to get the whole area, = π 3 + π + 4π 3 3 = 4π 3 3 8π 3 3 4.93 If I had to guess, I would say that region occupies about a third of the circle of radius, which has area 4π.6, a third of which would be about 4.9, so our answer seems reasonable.
(b) Let s graph the area we re trying to determine: For some angles (the first quadrant), this is an area between curves, but the inside curve never makes it into the second quadrant, so we have to treat that area differently. Look at it this way: In the blue region, the curve r = + cos θ is the outside curve and the curve r = cos θ is the inside curve. In the red region, we only have the r = + cos θ curve. We should derive the angle at which the second curve reaches so we can be sure what our integral bounds should be. cos θ is zero where cos θ is zero, which happens at θ = π. Let s find the area of the top half of the graph, and then double it, using the symmetry of the graph to reduce our work. I ll write the area between curves integral for the first quadrant, and then the area inside the curve that goes into the second quadrant as separate parts: / ( + cos θ) ( cos θ) dθ + π/ ( + cos θ) dθ
Since I have all these factors of floating around, let me multiply both sides by to clean them up: / ( + cos θ) ( cos θ) dθ + ( + cos θ) dθ π/ Expanding squares and combining like terms, / + cos θ 3 cos θ dθ + + cos θ + cos θ dθ π/ Applying the half-angle identity to the cosine square terms: / Doing a little clean-up, Evaluating: / + cos θ 3 + cos(θ) + cos θ 3 cos(θ) dθ + dθ + + cos θ + + cos(θ) dθ π/ π/ 3 + cos θ + cos(θ) dθ [ θ + sin θ 34 ] π/ [ 3 sin(θ) + θ + sin θ + ] π 4 sin(θ) = ( π4 + sin π 34 sin π + sin 34 ) sin ( 3 + π + sin π + 4 sin π 3π 4 sin π ) 4 sin π = ( π ) ( 3π 4 + + 3π ) 4 = π This area is harder to estimate using circular areas, but it does not seem wildly unreasonable, given the area of the unit circle (pictured on the graph) is π. π/