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Unformatted text preview: Math 2220 Problem Set 8 Solutions Spring 2010 These are problems based on sections 3.1 and 3.2 of the book. 1. Consider the curve parametrized by c ( t ) = (2 cos t, 3 sin t ) (a) Verify that this curve is an ellipse by finding an equation in x and y that is satisfied by all points on the curve. Solution. We have that sin t = x ( t ) / 2 and cos t = y ( t ) / 3. The equation sin 2 t + cos 2 t = 1 gives parenleftbigg x ( t ) 2 parenrightbigg 2 + parenleftbigg y ( t ) 3 parenrightbigg 2 = 1 , therefore the points of the curve satisfy the equation 9 x 2 + 4 y 2 = 36 which defines an ellipse. (b) At the point on the curve where t = / 3 find the velocity vector, the speed, and the tangent line (in parametrized form). Solution. The derivative is c ( t ) = ( 2 sin t, 3 cos t ). At t = / 3 we have c ( / 3) = ( 3 , 3 / 2) , which is the velocity of the particle moving along the curve at that moment. Its speed is given by the length of the vector c ( t ) which is | c ( t ) | = radicalbig 3 + 9 / 4 = 21 / 2. A parametrization of the tangent line is given by r ( u ) = c ( t ) + u c ( t ) = parenleftBigg 1 3 u, 3 3 + 3 u 2 parenrightBigg . (c) Find the points on the curve where the tangent line has slope 1. Solution. The tangent line has slope 1 when the tangent vector is perpendicular to the vector (1 , 1), i.e., when 2 sin t + 3 cos t = 0 or tan t = 3 / 2. The coordinates of the two points on the curve which satisfy this condition are (2 cos t, 3 sin t ) , where t = tan 1 3 / 2 or t = tan 1 3 / 2 + i.e., parenleftbigg 4 13 , 9 13 parenrightbigg and parenleftbigg 4 13 , 9 13 parenrightbigg ....
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