UNIVERSITY OF TORONTO DEPARTMENT OF MATHEMATICS MAT 237 Y - MULTIVARIABLE CALCULUS FALL-WINTER 2005-06
TEST #1. SOLUTIONS
1. a) Compute the area of the region bounded by the curve x = 2 + t 3 , y = 1 + t + 5 t 2 and the lines x = 2 , x = 3 and y = 0 . Soln. At x = 2 , t = 0 and at x = 3 , t = 1 . Then, A = = = = . b) Determine whether the curve x = 2 t – t 2 , y = 2 + 3 t – t 2 is concave upward or concave downward at the point ( 0 , 4 ) . Soln. , then . At ( 0 , 4 ) , t = 2 and . The curve is concave downward at ( 0 , 4 ) .
2. Compute the length of the polar curve r = sin 3 (q / 3 ) , 0 £ q £ 3 p / 4 . Soln. = = = = = = .
3. Suppose that a and b are two non-zero vectors such that and . Compute . Soln. Let q denote the angle between the vectors a and b , then , and from the first given condition we obtain (eq. 1) . From the second given condition we obtain or (eq. 2) . Adding (eq. 1) to the square of (eq. 2) , we obtain . Therefore, .
4. Find the distance from the point ( 2 , 0 , – 1 ) to the line of intersection of the planes 2 x + y = 1 and y + z = 2 . Soln. Let L denote the line of intersection of the given planes and let P ( 2 , 0 , – 1 ) . The point A ( 1 , – 1 , 3 ) belongs to the line L and v = ( 2 , 1 , 0 ) ´ ( 0 , 1 , 1 ) = ( 1 , – 2 , 2 ) is a vector director of the line L . Now, = ( 1 , 1 , – 4 ) , and . Then, the distance d from the point P to the line L is .
5. Let L be line that is tangent to the curve x = 1 + t 2 , y = 1 – t , z = 7 + t – t 2 , at the point ( 5 , 3 , 1 ) . Find the coordinates of the point of intersection of the line L and the plane x – y = 8 . Soln. ( x ¢ , y ¢ , z ¢ ) = ( 2 t , – 1 , 1 – 2 t ) . At ( 5 , 3 , 1 ) , t = – 2 and the vector v = ( – 4 , – 1 , 5 ) is tangent to the curve at ( 5 , 3 , 1 ) . Then, x = 5 – 4 l , y = 3 – l and z = 1 + 5 l are parametric equations of the corresponding tangent line. At the intersection with the plane x – y = 8 , ( 5 – 4 l ) – ( 3 – l ) = 8 and l = – 2 . The coordinates of the intersection point are ( 13 , 5 , – 9 ) .
6. Suppose that r ( t ) is a vector function such that r ¢ ( t ) exists for all real numbers t . a) Let f ( t ) =½r ( t )½5 . Determine the values of the constants m and n , if any, for which f ¢ ( t ) = m½r ( t )½n . Soln. , then = 5½r ( t )½ 3 . Therefore, m = 5 and n = 3 . b) Let v ( t ) = t 3 r ( t ) . Determine the polynomial function p ( t ) , if any, for which ( r ( t ) + r ¢ ( t ) ) ´ v ¢ ( t ) = p ( t ) ( r ( t ) ´ r ¢ ( t ) ) . Soln. v ¢ ( t ) = 3 t 2 r ( t ) + t 3 r ¢ ( t ) , then r ( t ) ´ v ¢ ( t ) = t 3 ( r ( t ) ´ r ¢ ( t ) ) and r ¢ ( t ) ´ v ¢ ( t ) = 3 t 2 ( r ¢ ( t ) ´ r ( t ) ) . Therefore, ( r ( t ) + r ¢ ( t ) ) ´ v ¢ ( t ) = ( t 3 – 3 t 2 ) ( r ( t ) ´ r ¢ ( t ) ) . That is, p ( t ) = ( t 3 – 3 t 2 ) .
7. Let . a) Show that the function f is not continuous at ( 0 , 0 ) . Soln. Approaching ( 0 , 0 ) along the line ( t , t ) we obtain, for t ¹ 0 , ,
and f is not continuous at ( 0 , 0 ) . b) Compute f x ( 0 , 0 ) and f y ( 0 , 0 ) , if they exist . Soln. Notice that g ( x ) = f ( x , 0 ) = 1 for all values of x . Then, g ¢ ( x ) = f x ( x , 0 ) = 0 for all values of x , and f x ( 0 , 0 ) = 0 . Similarly, h ( y ) = f ( 0 , y ) = ( 1 – y ) 2 , h ¢ ( y ) = f y ( 0 , y ) = – 2 ( 1 – y ) , and f y ( 0 , 0 ) = – 2 .
8. a) Prove that for any two sets A and B in R n , int ( A Ç B ) = int ( A ) Ç int ( B ) . Is it also true that for any two sets A and B in R n , int ( A È B ) = int ( A ) È int ( B ) ? Soln. We will use the fact that the interior of any set S is just the largest open set contained in S . That is, for any open set U , if U Í S then U Í int ( S ) . Now, int ( A Ç B ) Í ( A Ç B ) , therefore int ( A Ç B ) Í A and int ( A Ç B ) Í B . But int ( A Ç B ) is an open set, then int ( A Ç B ) Í int ( A ) and int ( A Ç B ) Í int ( B ) . Therefore int ( A Ç B ) Í ( int ( A ) Ç int ( B ) ) (cond. 1) . Also, int ( A ) Í A and int ( B ) Í B , then ( int ( A ) Ç int ( B ) ) Í ( A Ç B ) . But int ( A ) Ç int ( B ) is an open set because both int ( A ) and int ( B ) are open. Therefore ( int ( A ) Ç int ( B ) ) Í int ( A Ç B ) (cond. 2) . From (cond. 1) and (cond. 2) , we obtain int ( A Ç B ) = int ( A ) Ç int ( B ) . The property does not hold if the intersections are replaced by unions. Here is a counterexample. Take A = [ 0 , 1 ] and B = [ 1 , 2 ] . Then A È B = [ 0 , 2 ] , int ( A ) = ( 0 , 1 ) , int ( B ) = ( 1 , 2 ) and int ( A È B ) = ( 0 , 2 ) ¹ ( 0 , 1 ) È ( 1 , 2 ) = int ( A ) È int ( B ) . b) Prove or disprove the following proposition: If f : C ® R n is continuous, where C Í R m and C is closed, then f ( C ) Í R n is also closed. Soln. The given proposition is false. The following is a simple counterexample. Take C = [ 1 , ¥ ) Í R and f : C ® R defined as f ( x ) = 1 / x . Notice that C is closed and f is continuous but f ( C ) = ( 0 , 1 ] , which is not a closed set in R .
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