MATH20C_HW2Q56

MATH20C_HW2Q56 - x y z =(0,0,0 into T and check if the...

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56. A fighter plane, which can only shoot bullets straight ahead, travels along the path ( ) 2 3 5 ,21 ,3 / 27 t t t t = - - - r . Show that there is precisely one time t at which the pilot can hit a target located at the origin. Since the bullets will only fly along the direction in which the plane is heading, we need to know the direction of the plane, which is parallel to the velocity vector. The velocity vector is: ( ) 2 ' 1, 2 , / 9 t t t = - - - r If the bullet shot by the plane at time t hits the origin, then the tangential line of the path r ( t ) should join the point where the plane is situated at t and the origin. The co-ordinate P of the plane at t is given by: ( ) 2 3 : 5 ,21 ,3 / 27 P t t t - - - and therefore, the tangential line T at P is: ( ) ( ) ( ) 2 3 2 21 3 / 27 5 : 1 2 / 9 y t z t x t T t t - - - - - - = = - - - Since we now want to know whether the origin is on the tangential line or not, we substitute
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Unformatted text preview: ( x , y , z ) = (0,0,0) into T and check if the equation yields any feasible solution of t . First, check x and y : ( ) ( ) ( ) ( )( ) 2 2 2 21 5 1 2 2 5 21 10 21 3 7 t t t t t t t t t t----=---=--+ =--= which means t can be 3 or 7. Next, check if any of these two possible solutions also conforms with y-z equation Substitute t = 3, L.H.S. : (21-3 2 )/2(3) = 2 R.H.S. : 9(3-3 3 /27)/3 2 = 2 = L.H.S. So t = 3 is a solution. Substitute t = 7: L.H.S. : (21-7 2 )/2(7) = -2 R.H.S. : 9(3-7 3 /27)/7 2 = -1.78 ≠ L.H.S. which means t = 7 is not a valid solution. That implies the pilot can hit the target at the origin only when he/she pulls the trigger at t = 3....
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