Math 323 Exam 1 Practice Problem Solutions

Transcription

1 Math Exam Practice Problem Solutions. For each of the following curves, first find an equation in x and y whose graph contains the points on the curve. Then sketch the graph of C, indicating its orientation. (a) x t + and y t Equation: y x [Notice x t ] t> t< 7 (b) x t and y t Equation: y ±x [Notice t ± x ±x ] t> t< (c) x e t, y e t Equation: y x [Notice y (e t ) ]

2 t> t t< (d) x cost, y sint Equation: x + y 9 [Use: sin t+cos t ] (e) x cost, y sint Equation: x y [Use: cos(t) sin t]. Given the parametric curve: x t +, y t + (a) Find the equation of the tangent line to this parametric curve corresponding when t. Recall that dy dx dy dt dx dt 9t t 9 t, which when t gives m 9. Also, when t, the point on the curve is (,). Therefore the equation for the tangent line to the graph of this parameterized curve is given by: y 9 (x ). (b) Compute d y dx and use this to determine the concavity of the curve when t. d y dx d dx (y ) dy dt dx dt t. 9 t 9 9 t, which when t gives, so the graph of the parametric curve is concave up when

3 . (a) Graph the parametric curve given by x cos t, y sin t. (b) Prove that the graph of this parametric curve satisfies the equation x +y. Notice that x cos t and y sin t. Thus x +y.

4 . Find the arc length of the curve given by: (a) x t, y t, t Notice that x (t) and y (t) t. Therefore, L +t dt (+t) (b) x sint cost, y sint+cost, t Notice that x (t) cost+sint and y (t) cost sint Then L (cost+sint) +(cost sint) dt dt t [ ] [ ] () (9) 7 ( 7) cos t+sintcost sin t+cos t sintcost+sin tdt. Find the surface area of the solid obtained by revolving the parameterized curve given by revolving x t, y t+ for t about the y-axis. S x ds t (t ) +() dt t 9t + dt If we let u 9t +, then du t dx, or du t dx. ( ) Then we have S u du u ] [ Express the following polar equations in rectangular coordinates: (a) r cosθ Multiplying both sides by r gives r rcosθ. Then x +y x, or x +x+y Completing the square, (x+ ) +y (b) r sin(θ) By the double angle identity for sinθ, r sinθcosθ. Then r rsinθrcosθ, or (x +y ) xy Then (x +y ) x y 7. Express the following rectangular equations in polar coordinates: (a) xy rcosθrsinθ, or r sinθcosθ Then r secθcscθ (b) x y r cos θ r sin θ, or r (cos θ sin θ) By the double angle identity for cosθ, r cos(θ). Then r cos(θ), or r sec(θ). 8. Find the equation for a circle with center (, ) and passing through the origin in both rectangular and polar coordinates. Rectangular: x +(y +) Expanding, x +y +8y +, or x +y 8y Then r 8rsinθ, or r 8sinθ

5 9. Graph each of the following polar equations: (a) r sinθ K. K... K. K. K. K. (b) r +cosθ K K K K K K (c) r cos(θ) K K K (d) r sin(θ) K. K. K.... K (e) r θ K K K K K

6 . Find the area of each of the following polar regions: (a) the region bounded by the polar graph r +cosθ K. K. A (+cosθ) dθ +cosθ +cos θ dθ +cosθ + + cos(θ) dθ θ +sinθ + sin(θ) (b) the region bounded by one loop of the curve r sin(θ) K. K. K.... K Notice that sin(θ) when θ k or when θ k. Then the first loop is traced on when θ. Therefore, A (sin(θ)) dθ sin (θ) dθ cos(θ) dθ [ θ sin(θ) (c) the region inside r +sinθ and outside r ] K K K K K K K K First notice that +sinθ when sinθ, or sinθ. That is, when θ +k or +k. Then, using symmetry, A ((+sinθ) () ) dθ 7+sinθ +( cos(θ)) dθ θ cosθ sin(θ) ( ) ( +sinθ cos(θ) dθ () ) 9+sinθ +sin dθ

7 (d) the region inside both r cosθ and r sinθ K K Notice that cosθ sinθ when tanθ, or when θ +k. Also notice that sinθ is the boundary curve for θ while cosθ is the boundary curve for θ. Then A (sinθ) dθ + cos(θ) dθ + [( ) ] () + (cosθ) dθ +cos(θ) dθ θ sin(θ) [ ( + ) ( + )]. Find the arc length of the polar curve r sinθ sin θ dθ + +θ + sin(θ) cos θ dθ K K K K K K Notice that r (θ) cosθ and recall that ds r + ( ) dr dθ Setting up the integral is fairly routine: Using symmetry, L ( sinθ) +( cosθ) dθ 8sinθ +sin θ +cos θ dθ 8 8sinθ dθ, which, by the co-function identity for sinθ and cosθ 8 8cos( θ) dθ Which, using the half angle identity, cos( θ) ( ) cos( θ) dθ sin ( θ) dθ 8 cos( θ) cos( θ). Given the vectors a, and b,, find: (a) a+ b, (b) a b,,, 7 (c) a b 9,, 7, sin( θ) dθ

8 (d) a unit vector in the direction of a b a b,, and,. Hence our unit vector is:, (e) a vector with magnitude in the direction of a+ b a+ b, +, 8,, and 8, Hence the desired vector is: 8 7, 7 8 7, 8 7. Suppose that the thrust of an airplane s engine produces a speed on mph in still air and wind velocity is given by,. Find the direction the plane should head in order to fly due north. Also find the speed at which the plane will travel this course. Let v x,y be the velocity vector of the airplane and w, be the wind vector. Since the airplane ends up flying due north, we know that v + w,s. Therefore, v,s. Let y s Now, v +y +y, so, + y. Thus 9, y, so y Notice that arctan , so the plane should head.87 East of North. Finally, the speed of the airplane is s miles per hour.. The water from a fire hose exerts a force of lbs on the person holding the hose. The nozzle of the hose weighs lbs. Find the force required to hold the hose horizontal and find the angle to the horizontal that this force must be applied. Notice that the total force acting on the hose (from the water presure combined with gravity) is h,. Then the force required to keep the hose level acts equal and opposite these forces: v, Hence v + lbs of force. The angle above the horizontal at which this force must be applied is given by: θ arctan( ).7. Given P(,,) and Q(,, ) (a) Plot P and Q. (I ll let you figure out this one). (b) Find d(p,q) () +() +( ) ++. (c) Find the midpoint of the line segment between P and Q. ( ) + M, +, +( ) (,, ) (d) Find the equation for the sphere centered at P and passing through Q. (x+) +y +(z ). Given the vectors v i j +k and w i+j k (a) Find v + w i+j +k (b) Find v w (i j +k) (i+9j k) 8i j +9k (c) Find w v i+j k ++ (d) Find a unit vector in the same direction as v. Since v i j +k ++9, the unit vector we are looking for is: i j + k (e) Find a vector with magnitude in the opposite direction as w. Since w i+j k +9+, the vector we are looking for is: i j + k

9 7. Find the force needed to keep a helicopter stationary if the helicopter weighs lbs and a northeasterly wind exerts a force of lbs on the helicopter. F g,, is the force of gravity on the helicopter (in lbs). F w cos( ),sin( ), 7,7, is the force due to the wind (in lbs). Then,toopposetheseforcesandkeepthehelicopterstationary,thehelicoptermustsupply: Fh 7, 7,. That is, 7 lbs West, 7 lbs South, and lbs Up. The magnitude of the force that must be generated by the helicopter is: (7 ) +(7 ) +(),+,+,,,,.87