Questions and Answers #3 Negative Numbers & Coordinate Plane

Questions and Answers Sheet 3 Negative Numbers & Coordinate Plane Question #1 Find the terminal point P(x, y) on the unit circle determined by the given value of t. 𝑡= 5𝜋 2 Answer: We must find the angle that is coterminal with the given angle 𝑡 = 5𝜋 2 that is within the interval [0,2𝜋). Coterminal angles are angles that are apart by any integer multiple of 2𝜋 . Notice that when we subtract 2𝜋 to t, we get the following angle: 5𝜋 𝜋 𝑡≡ − 2𝜋 = ∈ [0,2𝜋) 2 2 𝜋 Looking at the values of the standard points in the unit circle, the angle 𝑡 = corresponds to the terminal 2 point (0,1) on the unit citcle. Result: (0,1) Question #2 Show that the distance between the parallel planes 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑1 = 0 and 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑2 = 0 is 𝐷 = (𝑎2 |𝑑1−𝑑2| +𝑏 2 +𝑐 2 )1 /2 Answer: The distance of the point (𝑥1 , 𝑦1 , 𝑧1 ) from the plane ax+by+cz+d=0 is given by 𝐷= 𝑎𝑥1 +𝑏𝑦1 +𝑐𝑧1 +𝑑 √𝑎2 +𝑏 2 +𝑐 2 If (𝑥1 , 𝑦1 , 𝑧1 ) is a point on the first plane, then its distance from the second plane, then its distance from the second plane is given by 𝐷= |𝑎𝑥1 +𝑏𝑦1 +𝑐𝑧1 +𝑑2 | √𝑎2 +𝑏 2 +𝑐 2 But (𝑥1 , 𝑦1 , 𝑧1 ) is a point on the first plane, this means that we can write 𝑎𝑥1 + 𝑏𝑦1 + 𝑐𝑧1 + 𝑑1 = 0 Subtract 𝑑1 from both sides 𝑎𝑥1 + 𝑏𝑦1 + 𝑐𝑧1 = −𝑑1 Substitute this in 𝐷= |−𝑑1 +𝑑2 | √𝑎2 +𝑏 2 +𝑐 2 Hence proved. Question #3 Find the length of the curve. 𝑟(𝑡) = 8𝑖 + 𝑡 2 𝑗 + 𝑡 3 𝑘, 0 ≤ 𝑡 ≤ 1 Answer: Given: The curve 𝑟(𝑡) = 8𝑖 + 𝑡 2 𝑗 + 𝑡 3 𝑘 for 0 ≤ 𝑡 ≤ 1 To find: The length of the curve. 𝑏 Answer: The arc length is given by ||𝑟(𝑡)|| = ∫𝑎 √(𝑥 ′ )2 + (𝑦 ′ )2 + (𝑧 ′ )2 dt where 𝑟(𝑡) = 𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 obtain the value as 𝑑 𝑥′ = (8) = 0 𝑑 𝑦 ′ = (𝑡 2 ) = 2𝑡 𝑑𝑡 𝑑 𝑧 ′ = (𝑡 3 ) = 3𝑡 2 𝑑𝑡 𝑑𝑡 Substitute the values in the formula. The length for the interval 0 ≤ 𝑡 ≤ 1 becomes 1 ||𝑟(𝑡)|| = ∫ √(0)2 + (2𝑡)2 + (3𝑡 2 )2 𝑑𝑡 0 1 = ∫0 √4𝑡 2 + 9𝑡 4 𝑑𝑡 1 ∫0 𝑡√4 + 9𝑡 2 𝑑𝑡 Take 𝑠 = 4 + 9𝑡 2 , differentiating with respect to t gives ds=18tdt 1 𝑡𝑑𝑡 = 𝑑𝑠 18 At 𝑡 = 0, 𝑠 = 4 + 9(0)2 = 4 At 𝑡 = 1, 𝑠 = 4 + 9(1)2 = 13 Thus, the integral with the above substion becomes 1 ||𝑟(𝑡)|| = ∫0 𝑡√4 + 9𝑡 2 𝑑𝑡 = = = 1 13 ∫ 18 4 1 2 [ 𝑠 ] 18 3 1 27 √𝑠𝑑𝑠 3 13 2 3 2 4 3 (13 − 42 ) 1 That is, ||𝑟(𝑡)|| = 27 (13√13 − 8) Question #4 Find the area of the surface generated when the given curve is revolved about the x-axis. 𝑦= 𝑥3 6 + 1 2𝑥 , for 1 2 ≤𝑥≤1 Answer: We have, 𝑦= 𝑥3 6 + 1 2𝑥 1 ; 𝑥 ∈ [ , 1] Here, 𝑓(𝑥) = 2 𝑥3 6 + 1 2𝑥 4 ⇒ 𝑓 ′ (𝑥) = 𝑥2 2 − 1 2𝑥 2 𝑥 1 1 𝑥2 1 1+𝑓 = 1 + + 4 − = ( + 2) 4 4𝑥 2 2 2𝑥 1 𝑥3 1 𝑥2 1 ∴ 𝑆 = 2𝜋 ∫ ( + ) ( + 2 ) 𝑑𝑥 2𝑥 2 2𝑥 1/2 6 5 𝑥 𝑥 𝑥 1 ⇒ 𝑆 = 2𝜋 ∈ 𝑡 11/2 [ + + + 3 ] 𝑑𝑥 12 12 4 4𝑥 263 ⇒ 𝑆 = 2𝜋 [ ] 512 263 Answer 𝑆 = 2𝜋 [ ] 2 ′ (𝑥)2 512 Question #5 Find the curvature K of the curve at the point P. 𝑟(𝑡) = 𝑡𝑖 + 𝑡 2 𝑗 + 𝑡 3 /4𝑘, 𝑃(2,4,2) Answer: Given 𝑟(𝑡) = 𝑡𝑖 + 𝑡 2 𝑗 + 𝑡3 4 𝑘 an point P(2,4,2) ⇒ 𝑥(𝑡) = 𝑡, 𝑦(𝑡) = 𝑡 2 and 𝑧(𝑡) = 𝑡3 4 At P(2,4,2) 𝑥(𝑡) = 2 ⇒ 𝑡 = 2 Now, 3 𝑟 ′ (𝑡) = 𝑖 + 2𝑡𝑗 + 𝑡 2 𝑘 4 At t=2 3 𝑟 ′ (2) = 𝑖 + 2(2)𝑗 + (2)2 𝑘 = 𝑖 + 4𝑗 + 3𝑘 4 ⇒ ||𝑟 ′ (2)|| = √12 + 42 + 32 = √1 + 16 + 9 = √26 3 3 𝑟 ′′ (𝑡) = 2𝑗 + (2𝑡)𝑘 = 2𝑗 + 𝑡𝑘 4 3 2 𝐴𝑡𝑡 = 2, 𝑟 ′′ (2) = 2𝑗 + (2)𝑘 = 2𝑗 + 3𝑘 2 𝑟 ′ (2) × 𝑟 ′′ (2) = (𝑖 + 4𝑗 + 3𝑘) × (2𝑗 + 3𝑘) = 6𝑖 − 3𝑗 + 2𝑘 ⇒ ||𝑟 ′ (2) × 𝑟 ′′ (2)|| = √62 + (−3)2 + 22 = √36 + 9 + 4 = √49 = 7 Curvature of a curve at a given point is given by || r (t )  r (t ) || r (t ) ||3 || r (t )  r (t ) 7 7 ⇒ at t = 2, K = = = 3 3 || r (t ) || ( 26) 26 26 K= Question #6 Find a Cartesian equation for the curve and identify it. 𝑟 2 cos 2 𝜃 = 1 Answer: Since cos 2 𝜃 = 2 cos 2 𝜃 − 1, then 𝑟 2 𝑐𝑠𝑜2𝜃 = 2𝑟 2 cos 2 𝜃 − 𝑟 2 = 2(𝑟 cos 𝜃)2 − 𝑟 2 = 2𝑥 2 − (𝑥 2 + 𝑦 2 ) = 𝑥 2 − 𝑦2 so your curve is 𝑥 2 − 𝑦 2 = 1 which is known to be a hyperbola having the lines 𝑦 = ±𝑥 as asymptotes. Question #7 Determine the set of points at which the function is continuous. 𝑓(𝑥, 𝑦) = 𝑥 2 𝑦 3 /2𝑥 2 + 𝑦 2 if (x,y) not equal to (0,0) and 1 if (x,y)=(0,0) Answer: On 𝑅𝟚 without the origin, f is a rational function, and so is continuous at every point in its domain. Since it is defined on this entire domain, we see that f is continuous at every point except the origin, we must check whether lim 𝑓(𝑥, 𝑦) = 𝑓(0,0) (𝑥,𝑦)→(0,0) Checking along the line y=0 , we see that lim 𝑥 2 (0)3 𝑥→0 2𝑥 2 +(0)2 = lim 0 = 0 𝑥→0 But this implies that if f has a limit as (x,y) tends to (0,0), it must be 0, which is not the value of f(0,0). Thus we see that in this cse, f(x,y) is not continuous at the origin. Question #8 Find the planes tangent to the following surfaces at the indicated points: 1 a) 𝑥 2 + 2𝑦 2 + 3𝑥𝑧 = 10, at the point (1,2, ) 3 b) 𝑦 2 − 𝑥 2 = 3 at the point (1,2,8) c) xyz=1 , at the point (1,1,1) Answer: Here we set 𝑓(𝑥, 𝑦, 𝑧) = 𝑥 2 + 2𝑦 2 + 3𝑥𝑧, 𝑥0 = (1,2,1/3) and k=10 . Note that 𝑓(𝑥0 ) = 𝑘 Now we can calculate: ∇𝑓(𝑥) = (2𝑥 + 3𝑧, 4𝑦, 3𝑥) ⇒ ∇𝑓(1,2,1/3) = (3,8,3) Using equation we easily get: 1 1 3 3 0 = (3,8,3) ⋅ (𝑥 − 1, 𝑦 − 2, 𝑧 − ) = 3(𝑥 − 1) + 8(𝑦 − 2) + 3 (𝑧 − ) Thus, the tangent plane equation is: 3x+8y+3z=20 Question #9 At what point does the curve have maximum curvature? 𝑦 = 9 ln(𝑥) What happens to the curvature as 𝑥 → ∞ ? Answer: |𝑦 ′′ | 𝑘 = (1+(𝑦′ )2)3/2 9 9 | 2 x x2 = = 3 9 (1 + ( ) 2 )3/2 [1 + 812 ]2 x x |− = 9𝑥(𝑥 2 + 81)−3/2 To maximize the numerature, we find k'=0 𝑘 ′ = 9(𝑥 2 + 81)−3/2 + 9𝑥 ⋅ (−3𝑥)(𝑥 2 + 81)−5/2 = 9(𝑥 2 + 81)−5/2 [(𝑥 2 + 81) − 3𝑥 2 ] 9(81−2𝑥 2 ) = (𝑥 2 +81)5/2 → Setting k'=0 yields 𝑥 = ± Since k'<0 per 𝑥 < − 9 √2 9 √2 = ±6.3639 and k'>0 per 𝑥 ≻ 9 √2 , minima at 𝑥 = −9/√2 = 6.3639 Since k'>0 per 𝑥 < 9/√2 and k'<0 per 𝑥 > 9/√2 maxima at 𝑥 = 9/√2 → +6.3639 9𝑥 lim 𝑥→∞ (𝑥 2 +81)3/2 =0 → curvature tendo to 0 per lange x Question #10 𝜋 Write the equation of the line tangent to 𝑓(𝑥) = tan 𝑥 + 3 at ( , 4) 4 Find all values of x such that the tangent line to 𝑓(𝑥) = (𝑥 2 − 9)2 is horizontal. Answer: Given 𝑓(𝑥) = tan 𝑥 + 3 𝑦 = 𝑓(𝑥) = tan 𝑥 + 3 derivative f(x) with repect to 'x' 𝑑 𝑑 𝑓(𝑥) = (tan 𝑥 + 3) 𝑑𝑥 ′ (𝑥) 𝑓 𝑑𝑦 𝑑𝑥 𝑑𝑥 = sec 2 𝑥 + 0 = sec 2 𝑥 Slope ′𝑚′ = = sec 2 (𝜋/4) 2 = (√2) =2 𝑑𝑦 𝑑𝑧 /𝑥 = 𝜋 4 = sec 2 𝜋/4 Equation of tangent line is (𝑦 − 𝑦1 ) = 𝑚(𝑥 − 𝑥1 ) where (𝑥1 , 𝑦1 ) = (𝜋/4,4) and m=2 ⇒ (𝑦 − 4) = 2(𝑥 − 𝜋/4) 𝜋 𝑦 = 2𝑥 − + 4 𝑦 = 2𝑥 − 2 𝜋+8 2