Questions and Answers #1 Parametric Equations

Questions and Answers Sheet 1 Parametric Equations Question #1 Find an equation of the tangent plane to the given parametric surface at the specified point. 𝑥 = 𝑢2 + 1, 𝑦 = 𝑣 3 + 1, 𝑧 = 𝑢 + 𝑣; (5,2,3) Answer: We can write 𝑟(𝑢, 𝑣) = (𝑢2 + 1)𝑖 + (𝑣 3 + 1)𝑗 + (𝑢 + 𝑣)𝑘 We first compute the tangent vectors 𝑟𝑢 = 𝑟𝑣 = 𝜕𝑥 𝜕𝑢 𝜕𝑥 𝜕𝑣 𝑖+ 𝑖+ 𝜕𝑦 𝜕𝑢 𝜕𝑦 𝜕𝑣 𝑗+ 𝑗+ 𝜕𝑧 𝜕𝑢 𝜕𝑧 𝜕𝑣 𝑘 = 2𝑢𝑖 + 0𝑗 + 𝑘 𝑘 = 0𝑖 + 3𝑣 2 𝑗 + 𝑘 Thus the normal vector to the tangent plane is i j ru  rv = 2u 0 0 3v 2 k 1 1 Expand along Row 1 i 0 1 2u 1 2u 0 −j +k 2 3v 1 0 1 0 3v 2 = −3𝑣 2 𝑖 − 2𝑢𝑗 + 6𝑢𝑣 2 𝑘 Notice that the point (5,2,3) corresponds to the parameter values u=2 and v=1, So the normal vector there is = −3(1)2 𝑖 − 2(2)𝑗 + 6(2)(1)2 𝑘 = −3𝑖 − 4𝑗 + 12𝑘 If a is the position vector of a point on the plane. And n is a vector normal to the plane. Then the equation of the plane is (𝑟 − 𝑎) ∗ 𝑛 = 0 (𝑥 − 5) ∗ (−3) + (𝑦 − 2) ∗ (−4) + (𝑧 − 3) ∗ (12) = 0 -3x+15-4y+8+12z-36=0 -3x-4y+12z-13=0 3x+4y-12z+13=0 Result: 3x+4y-12z+13=0 Question #2 The Cartesian coordinates of a point are given. (i) Find polar coordinates (𝑟, 𝜃) of the point, where r > 0 and 0 ≤ 𝜃 < 2𝜋. (ii) Find polar coordinates (𝑟, 𝜃) of the point, where r<0 and 0 ≤ 𝜃 < 2𝜋. (3√3, 3) Answer: Here given rectangular coordinate (𝑥, 𝑦) = (3√3, 3) which is in First Quadrant Now use the relation 𝑟 = √𝑥 2 + 𝑦 2 we need to find out value of r>0 2 𝑟 = √(3√3) + 32 = 6 The angle 𝜃 is given by 𝑦 3 𝑥 3√3 𝜃 = tan−1 ( ) = tan−1 ( ) 1 𝜃 = tan−1 ( ) √3 Since the point is in first quadrant and tan 𝜋 6 = 1 √3 we will consider 𝜃 = 𝜋 6 𝜋 Polar coordinate for r>0 is (𝑟, 𝜃) = (6, ) 6 Other representation of r<0 are 𝑃(𝑟, 𝜃) = 𝑃(𝑟, 𝜃 + (2𝑛 + 1)𝜋) 𝜋 7𝜋 6 6 (−6, + 𝜋) = (−6, ) add 𝜋 Result: 𝜋 7𝜋 6 6 (6, ) and (−6, ) Question #3 Change from rectangular to spherical coordinates. (Let 𝜌 ≥ 0, 0 ≤ 𝜃 ≤ 2𝜋, and 0 ≤ 𝜙 ≤ 𝜋.) a) (0, − 8, 0)(𝜌, 𝜃, 𝜙) =? b) (−1, 1, − 2)(𝜌, 𝜃, 𝜙) =? Answer: Step 1 a) Consider the rectangular coordinate (0, − 8, 0) The objective is to convert the rectangular coordinate to spherical coordinate (𝜌, 𝜃, 𝜙) Here, 𝜌 ≥ 0, 0 ≤ 𝜃 ≤ 2𝜋, 0 ≤ 𝜙 ≤ 𝜋 Consider the following statement: The point P in 𝑅𝟛 with rectangular coordinates (x,y,z) and spheerical coordinates (𝜌, 𝜃𝜙) The relation between rectangular coordinates (x,y,z) and spherical coordinates (𝜌, 𝜃𝜙) is, 𝑥 = 𝜌 sin 𝜙 cos 𝜃 𝑦 = 𝜌 sin 𝜙 sin 𝜃 𝑧 = 𝜌 cos 𝜙 Step 2 The rectangular coordinate is (𝑥, 𝑦, 𝑧) = (0, − 8, 0) That is, 𝑥 = 0, 𝑦 = −8, 𝑧 = 0. Thus, from the statement, 𝜌 = √𝑥 2 + 𝑦 2 + 𝑧 2 = √02 + (−8)2 + 02 {𝑥 = 0, 𝑦 = −8, 𝑧 = 0} = √64 =8 From the statement, 𝑧 = 𝜌 cos 𝜙 0 = 8 cos 𝜙 {𝑧 = 0, 𝜌 = 8} cos 𝜙 = 0 𝜋 𝜙= 2 {0 ≤ 𝜙 ≤ 𝜋} Step 3 From the statement, 𝑦 = 𝜌 sin 𝜙 sin 𝜃 𝑦 sin 𝜃 = 𝜌 sin 𝜙 Substitute 𝑦 = −8, 𝜌 = 8, 𝜙 = sin 𝜃 = =-1 −8 8 sin 𝜋 2 𝜋 2 {sin 𝜃= 𝜋 2 3𝜋 = 1} 2 𝜋 {0 ≤ 𝜃 ≤ } 2 Hence, the spherical coordinate of the point (0, − 8, 0) is (8, 3𝜋 𝜋 2 , ) 2 Step 4 Consider the rectangular coordinate (−1, 1, − 2) The objective is to convert the rectangular coordinate to spherical coordinate (𝜌, 𝜃, 𝜙). Consider the following statement: The point P in 𝑅𝟛 with rectangular coordinates (x,y,z) and spherical coordinates (𝜌, 𝜃, 𝜙). The relation between rectangular coordinates (x,y,z) and spherical coordinates (𝜌, 𝜃, 𝜙) is, 𝑥 = 𝜌 sin 𝜙 cos 𝜃 𝑦 = 𝜌 sin 𝜙 sin 𝜃 𝑧 = 𝜌 cos 𝜙 The rectangular coordinate is (𝑥, 𝑦, 𝑧) = (−1, 1, − 2) That is, x=-1, y=1, z=-2. Thus, from the statement, 𝜌 = √𝑥 2 + 𝑦 2 + 𝑧 2 = √(−1)2 + (1)2 + (−2)2 {𝑥 = −1, 𝑦 = 1, 𝑧 = −2} = √1 + 1 + 4 = √6 From the statement, 𝑧 = 𝜌 cos 𝜙 −2 = √6 cos 𝜙 −2 cos 𝜙 = √6 −2 −1 𝜙 = cos ( ) √6 𝜙 = 144.74∘ {0 ≤ 𝜙 ≤ 𝜋} From the statement, 𝑦 = 𝜌 sin 𝜙 sin 𝜃 𝑦 sin 𝜃 = 𝜌 sin 𝜙 Substitute 𝑦 = 1, 𝜌 = √6, 𝜙 = 144.74∘ 1 sin 𝜃 = √6(144.74∘ ) 1 = √6(0.577287) sin 𝜃 = 0.7071 𝜃 = sin−1 (0.7071) = 45∘ 𝜋 {0 ≤ 𝜃 ≤ } 2 Hence, the spherical coordinate of the point (-1, 1, -2) is (√6, 45∘ , 144.74∘ ) Question #4 Express the plane z=x in cylindrical and spherical coordinates. a) cylindrical 𝑧 = 𝑟 cos(𝜃) b) spherical coordinates 𝜃 = arcsin(cot(𝜙)) Answer: Step 1 The plane z=x goes through the line intersection of the planes 𝜋 x=0 and z=0 and makes a angle with those planes. 4 So, a point (x,y,z)=(x,y,x)=(x,y) is on the plane. a) Cylindrical coordinates: Let 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃 with 𝑟 ≥ 0 and 0 ≤ 𝜃 ≤ 2 𝜋 Then. z=x ⇒ 𝑧 = 𝑟 cos 𝜃. So, a point on the plane takes the form (𝑥, 𝑦, 𝑧) = (𝑟 cos 𝜃 , 𝑟 sin 𝜃 , 𝑟 cos 𝜃) Step 2 b) Spherical coordinates: Let 𝑥 = 𝜌 sin 𝜙 cos 𝜃 , 𝑦 = 𝜌 sin 𝜙 𝜃, 𝑧 = 𝜌 cos 𝜙 with 𝜌 ≥ 0, 0 ≤ 𝜃 ≤ 2𝜋 and 0 ≤ 𝜙 ≤ 𝜋 Then, z=x ⇒ 𝜌 cos 𝜙 = 𝜌 sin 𝜙 cos 𝜃 𝜌 cos 𝜙 ⇒ = cos 𝜃 𝜌 sin 𝜙 ⇒ cot 𝜙 = cos 𝜃 ⇒ 𝜃 = arccos(cot 𝜙) So, a point on the plane takes the form (𝑥, 𝑦, 𝑧) = (𝜌 sin 𝜙 cos 𝜃 , 𝜌 sin 𝜙 sin 𝜃 , 𝜌 cos 𝜙) = (𝜌 sin 𝜙 cos(arccos(cot 𝜙)) , 𝜌 sin 𝜙 sin(arccos(cot 𝜙)) , 𝜌 cos 𝜙) = (𝜌 sin 𝜙 cos(arccos(cot 𝜙)) , 𝜌 sin 𝜙 sin(arcsin(√1 − cot 2 𝜙)) , 𝜌 cos 𝜙) = (𝜌 sin 𝜙 cot 𝜙 , 𝜌 sin 𝜙√1 − cot 2 𝜙 , 𝜌 cos 𝜙) = (𝜌 cos 𝜙 , 𝜌 sin 𝜙√1 − cot 2 𝜙 , 𝜌 cos 𝜙) Question #5 Let P(x)=F(x)G(x) and 𝑄(𝑥) = 𝐹(𝑥) 𝐺(𝑥) , where F and G are the functions whose graphs are shown. a) Find P'(2) b) Find Q'(7) Answer: Step 1 a) P'(2) will be the derivative of P(x) evaluated at x=2. So first, we take a derivative. In this section, you should have learned product rule, so P'(x) will look like this: P'(x)=F(x)G'(x)+F'(x)G(x) Now we let x=2 P'(2)=F(2)G'(2)+F'(2)G(2) Read these values off the graph. We see that F(2)=3 (When its x-value is 2, the curve F has a y-value of 3) We see that G(2)=2 . F'(2) will be the slope of the curve F at 2. A tangent line at F(2) would be horizontal, so F'(2)=0 . 1 1 G(2) has a slope of , so 𝐺 ′ (2) = 2 2 Now we substitute these values into P'(2) to solve: 1 3 3 𝑃′ (2) = (3) ( ) + (0)(2) = ( ) + (0) = 2 2 2 Step 2 We're playing by the same rules as above-differentiate the equation, read the values off the graph, substitute the values into the function, and solve. Differentiate (Quotient Rule): 𝑄′ (𝑥) = 𝐺(𝑥)𝐹′ (𝑥)−𝐹(𝑥)𝐺 ′(𝑥) 𝐺(𝑥)2 Read the values off the graph: G(7)=1 2 𝐺 ′ (7) = − 3 F(7)=5 1 𝐹 ′ (7) = 4 Substitute: 𝑄′ (7) = 1 4 2 3 [1⋅( )−5⋅(− )] [12 ] 1 10 4 43 3 𝑄′ (7) = [( ) + ( )] 𝑄′ (7) = ( ) = 3.5833 12 Question #6 Use the given graph off over the interval (0, 6) to find the following. a) The open intervals on which f is increasing. (Enter your answer using interval notation.) b) The open intervals on which f is decreasing. (Enter your answer using interval notation.) c) The open intervals on which f is concave upward. (Enter your answer using interval notation.) d) The open intervals on which f is concave downward. (Enter your answer using interval notation.) e) The coordinates of the point of inflection. (x, y)= Answer: The correct answers are actually: a) (1, 3), (4, 6) b) (0, 1), (3, 4) c) (0, 2) d) (2, 4), (4, 6) e) (2, 3) Question #7 Match the parametric equations with the graphs labeledI-VI. Give reasons for your choices. (Do not use a graphing device). a) 𝑥 = 𝑡 4 − 𝑡 + 1, 𝑦 = 𝑡 2 b) 𝑥 = 𝑡 2 − 2𝑡, 𝑦 = √𝑡 c) 𝑥 = sin 2 𝑡, 𝑦 = sin(𝑡 + sin 2 𝑡) d) 𝑥 = cos 5 𝑡, 𝑦 = sin 2 𝑡 e) 𝑥 = 𝑡 + sin 4 𝑡, 𝑦 = 𝑡 2 + cos 3 𝑡 f) 𝑥 = sin 2𝑡 4+𝑡 2 ,𝑦 = cos 2𝑡 4+𝑡 2 Answer: Step 1 Consider the parametric equation, This equation's graph matches with that of (V). As t goes to infinity, both x and y go to infinity. The graph must also contain the point (1, 0) achieved at t=0 However, you know that this function should not have any oscillatory behavior. Therefore, the parametric equations match graph (V). Step 2 Consider the parametric equation, 𝑥 = 𝑡 2 − 2𝑡, 𝑦 = √𝑡 The equation's graph matches with that of (I). As f goes to infinity, both x and y go to infinity. The only graphs that have this behavior are (V), (I) and (IV). The graph must also contain the point (0, 0) achieved at t=0 Only graph (I) satisifies this requirement. Step 3 Consider the parametric equation, 𝑥 = sin 2 𝑡, 𝑦 = sin(𝑡 + sin 2 𝑡) This equation's graph matches with that of (II). 𝜋 The graph must contain the point (0, 1) achieved at 𝑡 = . Graph (II) is the only graph that satisfies this. 2 Step 4 Consider the parametric equation, 𝑥 − cos 5 𝑡, 𝑦 = sin 2 𝑡 This equation's graph matches with that of (IV). The graph must contain the point (1, 0) achieved at t=0 . Graph (IV) is the only graph that satisfies this. Step 5 Consider the parametric equation. 𝑥 = 𝑡 + sin 4 𝑡, 𝑦 = 𝑡 2 + cos 3 𝑡 This equations graph matches with that of (IV). As t goes to infinity, both x and y go to infinity. The onli graph that have this behavior are (V), (I), and (IV). However, graph (IV) is the only one that shows any oscillatory behavior. Step 6 𝑥= sin 2𝑡 4+𝑡 2 ,𝑦 = cos 2𝑡 4+𝑡 2 This equation's graph matches with that of (III). The only graphs that exhibit oscillatory behavior are graphs (III), (II), (IV), and (VI). The graph must also satisfy the fact that the point (0, 0) cannot be a point on the graph. Graph (III) and (IV) satisfy this requirement. However, for graph (IV), as t goes to infinity, both x and y go to infinity. This is not the case for these equations. Question #8 Match the vector field F with the correct plot. F(x, y)=(x, -y) Answer: ( x, y ) (1, 0) (0, 1) (−1, 0) (0, − 1) (2, 0) (0, 2) f = ( x, − y ) (1, 0) (0, − 1) (−1, 0) (0, 1) (2, 0) (0, − 2) Clearly all points form first quadrant to all points of forte quadrant and vice versa. Similar all point of d Second quadrant to all point of third quadrant vice verse. Hence third picture Question #9 Data points (x, y) x 2 4 6 8 10 12 14 16 y 0.08 0.12 0.18 0.25 0.36 0.52 0.73 1.06 Draw scatter plots of (𝑥, ln 𝑦) and (ln 𝑥 , ln 𝑦) Answer: Step 1 In part (a), you already entered the x-coordinates under 𝐿1 and the y-coordinates under 𝐿2 in your graphing calculator. Enter 𝐿3 = ln(𝐿2 ) into your graphing calculator so 𝐿3 will fill in the values of 𝑙𝑛 y for you. Then go to STAT PLOT, choose Plot2, and change YList to 𝐿3 . Turn off Plot1 and turn on Plot2. Adjust the window to an appropriate viewing rectangle and then press GRAPH: Step 2 Enter 𝐿4 = ln(𝐿1 ) into your graphing calculator so 𝐿4 will fill in the values of 𝑙𝑛 x for you. Then go to STAT PLOT, choose Plot3, and change Xlist to 𝐿4 and Ylist to 𝐿3 . Turn off Plot2 and turn on Plot3. Adjust the window to an appropriate viewing rectangle and then press GRAPH: Question #10 𝑑𝑦 Find 𝑑2 𝑦 and 𝑑𝑥 2 . For which values of t is the curve concave upward? 𝑑𝑥 2 2 𝑥 = 𝑡 + 1, 𝑦 = 𝑡 + 𝑡 Answer: Using the chain rule, we can write 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 ⋅ 𝑑𝑡 𝑑𝑦/𝑑𝑡 = 𝑑𝑥 ⇒ (Equation 1) 𝑑𝑥/𝑑𝑡 It is given that 𝑥 = 𝑡 2 + 1 We can use the sum rule followed by the power rule to differentiate this 𝑑𝑥 = 2𝑡 2−1 + 0 = 2𝑡 𝑑𝑡 It is given that 𝑦 = 𝑡 2 + We can use the sum rule followed by the power rule to differentiate this 𝑑𝑦 = 2𝑡 2−1 + 1 = 2𝑡 + 1 𝑑𝑡 Substitute the expressions for dy/dt and dx/dt in (Equation 1), to get 𝑑𝑦 𝑑𝑥 = 2𝑡+1 2𝑡 =1+ 1 2𝑡 To find the concavity, we need the second derivative with respect r to x, so we will now differentiate both sides with respect to x 𝑑𝑦 𝑑 𝑑𝑥 𝑑𝑥 𝑑 𝑑𝑥 [ ]= 1 [1 + ] 2𝑡 In the right-hand side, we can use the chain rule to write 𝑑2𝑦 𝑑𝑥 2 𝑑2𝑦 𝑑𝑥 2 = 𝑑 𝑑𝑡 1 𝑑𝑡 2𝑡 1 𝑑𝑥 1 [1 + ] ⋅ = [0 − 2𝑡62 ]⋅ 𝑑𝑥/𝑑𝑡 In the red expression, replace dx/dt with 2t 𝑑2𝑦 𝑑𝑥 2 =− 1 2𝑡 2 ⋅ 1 2𝑡 =− 1 4𝑡 3 Remember that the graph is concave up/down depending on whether the second derivate is positive/negative respectively. Note that: (1) the number is never 0, so there are no sign changes that occur because of that. (2) the denominator is 0 when t = 0. This makes rational function discontinuous, so we will use that to divide the number line into intervals that need to be checked. To check thi sign on (−∞, 0), we will use t = -1 d2y 1 1 | t = −1 = − = 0 2 3 dx 4(−1) 4 To check the sign on (0, ∞), we will use t = 1 d2y 1 1 | t =1= − =− 0 2 3 dx 4(1) 4 Therefore, the graph is concave up on (−∞, 0) To identify all the intervals on which a function is positive/negative, follow the steps below: 1)Identify the points where the function is either zero or discontinuous 3) Use the result of (1) to divide the domain of the function into intervals. The function maintains the same sign for all numbers in an interval. 3) Pick a value from each interval to evaluate the function and identify its sign in that interval. 𝑑𝑦 𝑑𝑥 =1+ 1 𝑑2𝑦 2𝑡 𝑑𝑥 2 =− 1 4𝑡 3 The graph is concave up ob (−∞, 0)