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1/4 Evaluate the integral (1 + 3 cost) i + (3 sint) j+ (sect) k] at. - 1/4 The result is + j + K. (Type exact answers, using x and radicals as needed.)Solve the initial value problem for r as a vector function of t. dr Differential equation: dt = -7ti-t j - 5t k Initial condition: r(0) =61 + 8j + 9k r(t) = i + Dit kSolve the initial value problem for r as a vector function of t. dr Differential Equation: 1/2. = i+4e K dt (t + 1) + t+1 Initial condition: r(0) = k r(t) = KSolve the initial value problem for r as a vector function of t. Differential equation: = - 4k 2 dt dr Initial conditions: r(0) = 10k and = 10i + 10j dt t = 0 r(t) = + + KSolve the following initial value problem for r as a function of t. d Differential equation: = 2eli-e -fi + 12e 2k 2 dt Initial conditions: r(0) = 6i + 4j + 6k dr dt = - 2i + 5j It = 0 r(t) = KFind the parametrization for the line segment joining the points P(0,2,4) and Q(0, - 2,4). Draw coordinate axes and sketch the segment, indicating the direction of increasing t for the parametrization.Find the plane determined by the intersecting lines. L1 x= - 1+2t y =2+ 3t z =1-t 12 x=1-4s y = 1+2s z =2-2s Using a coefficient of - 1 for x, the equation of the plane is (Type an equation.)At time t= i}: a particle is located at the point {6.5.3}. It travels in a straight line to the point [4.9.1]: has speed 3 at [6.5.9] and constant acceleration 2i + 3j 8k. Find an equation for the position vector nit] of the particle at time t. The equation for the position vector r{t] of the particle at timet is rm: [D] H [ZED j+ [ED k. [Type exact answers. using radicals as needed.) A spring gun at ground level fires a golf ball at an angle of 45". The ball lands 10 m away. The acceleration due to gravity is g=9.8 m/s. What was the ball's initial speed? For the same initial speed, find the two firing angles that make the range 8 m. The initial speed of the ball is 9.90 m/sec. (Round to the nearest hundredth.) The two firing angles are degrees. (Round to the nearest tenth. Use a comma to separate answers as needed.)A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved along, a marble was fired into the air by a spring in the engine's smokestack. The marble, which continued to move with the same forward speed as the engine, rejoined the engine 1 sec after it was fired. The measure of the angle the marble's path made with the horizontal was 52". Use the information to find how high the marble went and how fast the engine was moving. The marble's maximum height was ft. (Type an exact answer. Type an integer or a decimal.)A projectile is fired with an initial speed of 550 m/sec at an angle of elevation of 30". Answer parts (a) through (d) below.a. When will the projectile strike? sec (Round to one decimal place as needed.) b. How far away will the projectile strike? m (Round to the nearest meter as needed.) c. How high overhead will the projectile be when it is 5 km downrange? m (Round to the nearest meter as needed.) d. What is the greatest height reached by the projectile? m (Round to the nearest meter as needed.)Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. 3/2 r(t) = 4ti + w/ N t k, Osts 33 4 The curve's unit tangent vector is K. /16 +t i + (0)j+ 16 +t (Type exact answers, using radicals as needed.) The length of the indicated portion of the curve is units. (Type an integer or a simplified fraction.)Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. r(t) = (cost)i+ ( sin >t) k, Osts-Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. r(t) = 6ti + 2t'j - 3t k 1sts2Match the equation with the graph. Include the directrix that corresponds to the focus at the origin. Label the vertices with appropriate polar coordinates. If the equation is an ellipse, label the center as well. 30 r= 12 -3 cos 0 Choose the correct graph below. O A. O B. O C. OD. 12 Ay Ay X X X X 121 42 42Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. r(t) = (5t sint+ 5 cost)i + (5t cost - 5 sint)j 1/2 sts2 The curve's unit tangent vector is ( cost ) i+ ( - sint )j + ( 0 ) k. The length of the indicated portion of the curve is units. (Simplify your answer.)Find the point on the curve r(t) = (5 sint)i + (-5 cost)j - 12tk at a distance 52x units along the curve from the point (0, -5,0) in the direction of increasing arc length. The point is (Type exact answers, using x as needed.)Find the arc length parameter along the given curve from the point where t=0 by evaluating the integral s(t) = _ v(t)|dt. Then find the length of the indicated portion of the curve r(t) = 8costi+ 8sint j + 5t k, where 0 Outs 4Find the arc length parameter along the curve from the point where t=0 by evaluating the integral s = _ lvldt. 0 Then find the length of the indicated portion of the curve. r(t) = (4e cost)i+ (4e sint)j-4e k, - In4sts0)Find the arc length parameter along the curve from the point where t = 0 by evaluating the integral s = . v(T) | dT. 0 Then find the length of the indicated portion of the curve. r(t) = (6 + 2t)i + (9 + 4t)j + (1 - 3t)k, - 1st0