Practice problems for Exam.. Given a =<,, > and b =<,, >. Find the area of the parallelogram with adjacent sides a and b. A = a b a ı j k b = = ı j + k = ı + 4 j 3 k Thus, A = 9. a b = () + (4) + ( 3) = 9. Find an equation of the line through the point (,, ) and perpendicular to the plane x + y + z = The line is parallel to the normal vector of the plane n =<,, >. Thus, symmetric equations of the line are: x = y = z + 3. Find the distance from the point (,, ) to the plane D = x + 3y + z = 7 + 3( ) + 7 () + (3) + () = 7. 4. Find an equation of the plane that passes through the point (, 3, ) and contains the line x = t, y = 4t, z = + t. The vector v =<, 4, > lies in the plane. Let P(, 3, ) and Q(,, ). The second vector that lies in the plane is the vector PQ =<, 3, >. Then the normal vector to the plane n = v ı j k PQ = 4 3 = ı 4 3 j + k 4 3 = ı + j 6 k Thus, an equation of the plane is (x + ) + (y + 3) 6(z ) = 5. Find parametric equations of the line of intersection of the planes z = x+y and x 5y z =. The direction vector for the line of intersection is v = n n, where n =<,, > is the normal vector for the first plane and n =<, 5, > is the normal vector for the second plane.
v = n n = ı j k 5 = ı 5 j + k 5 = ı( 5) j( + ) + k( 5 ) = 6 ı j 7 k. To find a point on the line of intersection, set one of the variables equal to a constant, say y =. Then the equations of the planes reduce to x z = and x z =. Solving this two equations gives x = z =. So a point on a line of intersection is (,, ). The parametric equations for the line are x = 6t y = t z = 7t 6. Are the lines x = + 4t, y = 3 + t, z = and x = 3 8s, y = s, z = parallel, skew or intersecting? If they intersect, find the point of intersection. The direction vector for the first line is v =< 4,, >, the second line is parallel to the vector v =< 8,, >. Since v = v, vectors v and v are parallel. Thus, the lines are parallel. 7. Identify and roughly sketch the following surfaces. Find traces in the planes x = k, y = k, z = k (a) 4x + 9y + 36z = 36 x 9 + y 4 + z = - ellipsoid z 3 x 3 y Traces in x = k: in y = k: in z = k: y 4( k) + z k 9 9 x = - ellipse 9( k) + z = - ellipse k 4 4 x 9( k ) + y 4( k ) = - ellipse (b) y = x + z An equation y = x + z defines the elliptic paraboloid with axis the y-axis.
3 z 3 4 x 4 3 4 5 y Traces in x = k: y = z + k - parabola in y = k: x + z = k - circle in z = k: y = x + k - parabola (c) 4z x y = An equation 4z x y = defines the hyperboloid on two sheets with axis the z-axis 3 z 3 5 x 5 6 4 y 4 6 Traces in x = k: z /4( k ) x k = - hyperbola z in y = k: /4( k ) y k = - hyperbola in z = k ( k > /): x + y = 4k - circle (d) x + z = An equation x + z = defines the elliptic cylinder with axis y-axis..5 z.5 x y Traces k k in x = k: z =, z = - lines in z = k: x = k, x = k - lines 8. Find lim t ( t + 3 ı + t t ( ) t t tan t lim + 3 ı + j + k t t t ) tant j + t k = lim t + 3 ı + lim j + lim t t t t t 4 ı + lim t t (t )(t + ) j + tan() k = ı + j + tan() k tant k = t 9. Find the unit tangent vector T(t) for the vector function r(t) =< t, sin t, 3 cost >. 3
The tangent vector r (t) =<, cost, 3 sin t >, r (t) = + 4 cos t + 9 sin t = + 4 cos t + 4 sin t + 5 sin t = 5 + 5 sin t. The unit tangent vector T(t) = r (t) = r (t) 5 + 5 sin t < t, cost, 3 sin t >. Evaluate ( t ı + te t j + ) t k dt = ( t ı + te t j + ) t k dt tdt ı + te t dt j + t dt k Thus, t 3/ 4 tdt = 3/ = 3 [43/ ] = 3 (8 ) = 4 3 te t dt = u = t u = v = e t v = e t = te t 4 + 4e 4 + e e 4 + e = 5e 4 + e t dt = 4 t = 4 + = 3 4 e t dt = 4e 4 + e e t 4 = ( t ı + te t j + ) t k dt = 4 3 ı + ( 5e 4 + e ) j + 3 4 k. Find the length of the curve given by the vector function r(t) = cos 3 t ı + sin 3 t j + cos(t) k, t π. r (t) = 3 cos t sin t ı + 3 sin t cos t j sin(t) k r (t) = 9 cos 4 t sin t + 9 sin 4 t cos t + 4 sin (t) Recall that sin(t) = sin t cost, then sin (t) = 4 sin t cos t and r (t) = sin t cos t(9 sin t + 9 cos t + 6) = sin t cost 5 = 5 sintcost = 5 sin(t) Then the length of the curve π/ 5 L = sin(t) dt = 5 4 cos(t) π/ = 5 4
. Find the curvature of the curve r(t) =< t 3, 3t, 6t >. κ(t) = r (t) r (t) r (t) 3 r (t) =< 6t, 6t, 6 >= 6 < t, t, >, r (t) = 6 + t + t 4 r (t) =< t, 6, > r (t) r (t) = ı j k 6t 6t 6 t 6 = ı 6t 6 6 j 6t 6 t + k 6t t 6t 6 = ı(36) j( 7t) + k( 36t + 7t ) =< 36, 7t, 36t >= 36 <, t, t > r (t) r (t) = 36 + 4t + t 4 Thus, κ(t) = 36 + 4t + t 4 (6 + t + t 4 ) 3 = + 4t + t 4 6( + t + t 4 ) 3/ 3. Find an equation of the normal plane to the curve r(t) =< t, t, lnt > at the point where t =. r() =<,, > The normal vector to the normal plane at the point where t = is r (). The equation of the normal plane is 4. Sketch the domain of the function r (t) =< t,, t > r () =<,, > (x ) + (y ) + (z ) = f(x, y) = x + y + ln(4 x y ) The expression for f makes sense if x + y and 4 x y >. Thus, the domain of the function f is D(f) = {(x, y) R x + y < 4} 5
5. Find the level curves of the function z = x y. An equation for the level curves is k = x y or y = x k. It defines the family of parabolas. 6. Find f xyz if f(x, y, z) = e xyz. f x = e xyz (xyz) x = yze xyz f xy = (yze xyz ) y = ze xyz + yze xyz (xyz) y = ze xyz + xyz e xyz f xyz = (ze xyz +xyz e xyz ) z = e xyz + ze xyz (xyz) z + xyze xyz + xyz e xyz (xyz) z = e xyz + xyze xyz + xyze xyz + x y z e xyz = ( + 3xyz + x y z )e xyz 7. The dimensions of a closed rectangular box are 8 cm, 6 cm, and 5 cm with a possible error of. cm in each dimension. Use differential to estimate the maximum error in surface area of the box. Let l, w, and h be the length, width, and height, respectively, of the box in centimeters. l = w = h =. The surface area of the box Thus, A(l, w, h) = (lw + lh + wh) A = A A A l + w + l w h h A A = w + h; (8, 6, 5) = l l A A = l + h; (8, 6, 5) = 6 w w A A = l + w; (8, 6, 5) = 8 h h A = ( + 6 + 8)(.) = 5cm 6
8. Find parametric equations of the normal line and an equation of the tangent plane to the surface x 3 + y 3 + z 3 = 5xyz at the point (,, ). Let F(x, y, z) = x 3 + y 3 + z 3 5xyz. Then F(x, y, z) =< 3x 5yz, 3y 5xz, 3z 5xy > F(,, ) =< 7, 7, 7 > The equation of the tangent plane at (,, ) is 7(x ) 7(y ) 7(z ) = The parametric equations of the normal line at (,, ) are x = + 7t, y = 7t, z = 7t 9. Given that w(x, y) = ln(3x + 5y) + x tan y, where x = s cott, y = s + sin t. Find w t. w t = w x x t + w y y t = ( 6 3x + 5y + ) csc t + ( 3x + 5y ) + y t. Let f(x, y, z) = ln(x+3y +6z). Find a unit vector in the direction in which f decreases most rapidly at the point P(,, ) and find the derivative (rate of change) of f in this direction. The function f decreases most rapidly in the direction of the vector f(,, ). f(x, y, z) = x + 3y + 6z, 3 x + 3y + 6z, 6 x + 3y + 6z f(,, ) = 3 + 6, 3 3 + 6, 6 =<, 3, 4 > 3 + 6 f(,, ) = 4 + 9 + 36 = 49 = 7 The unit vector in the direction of the vector f(,, ) is u = 7 <, 3, 4 >= 7, 3 7, 6 7 So, the function f decreases most rapidly in the direction of the vector u = 7, 3 7, 6 7 D u f(x, y, z) = f(x, y, z) ( u) = x + 3y + 6z, 3 x + 3y + 6z, 6 x + 3y + 6z 7, 3 7, 6 7 = 7 x + 3y + 6z 7
. Find z x and z y if xe y + yz + ze x = where F(x, y, z) = xe y + yz + ze x So z x = + ze x ey y + e x z x = x z, x = ey + ze x y = xey + z z = y + ex and z y = + z xey y + e. x. Find the local extrema/saddle points for z y = y z f(x, y) = x + y + xy + x + y We first locate the critical points: f x (x, y) = x + y + f y (x, y) = y + x + Setting these derivatives equal to zero, we get the following system: { x + y + = x + y + = Substitute y = x from the second equation into the first equation: x + ( x) + = x =, y = x = The critical point is (, ). Next we calculate the second partial derivatives: f xx (x, y) = 4, f xy (x, y) =, f yy (x, y) =. Then D(x, y) = 4 = 8 4 = 4 > Since D(x, y) = 4 > and f xx (x, y) = 4 >, the function f has a local minimum at the point (, ), f(, ) =. 3. Find the absolute maximum and minimum values of the function f(x, y) = x +xy +3y over the set D, where D is the closed triangular region with vertices (, ), (, ), and (, ). The set D is bounded by lines x =, y =, and x y = 8
First we find critical points for f: f x (x, y) = x + y = f y (x, y) = x + 6y = so the only critical point is (, ). f(, ) = Now we look at the values of f on the boundary of D. f(, ) = ( ) + ( )() + 3() = f(, ) = () + ()() + 3() = f(, ) = ( ) + ( )( ) + 3( ) = 7 If x =, then f(, y) = y + 3y f y (, y) = + 6y =, so y = /3. f(, /3) = /3 If y =, then f(x, ) = x + x +, f x (x, ) = x + = and x = f(, ) = If x y =, then y = x, and g(x) = f(x, y) = x +xy +3y 3 = x +x(x )+3(x ) = 6x 8x 3 g (x) = x 8 =, so x = 8/ = /3 and y = /3 = /3 f(/3, /3) = /3 Thus, the absolute maximum value of the function is f(, ) = 7 and the absolute minimum value is f(, ) =. 4. Use Lagrange multipliers to find the maximum and minimum values of the function f(x, y) = xy subject to the constraint 9x + y = 4. Let g(x, y) = 9x + y. f(x, y) =< y, x >, g(x, y) =< 8x, y >. Using Lagrange multipliers, we solve the equations 9
From the first equation λ = gives If y = 3x, then f ( ± ) 3, ± = 3 If y = 3x, then f ( ± ) 3, = 3 Thus, the minimum value is f 3. y = λ(8x) x = λ(y) 9x + y = y. Plugging the expressions for λ into the second equation, 8x x = y 8x y 9x = y 9x y = (3x y)(3x + y) = y = 3x or y = 3x 9x + y = 9x + (3x) = 8x = 4 x = ± and y = 3x = ± 3 9x + y = 9x + ( 3x) = 8x = 4 x = ± and y = 3x = 3 ( ± ) ( 3, = 3 ) and the maximum value f ± 3, ± =