MATH 004 Homework Solution Homework 8 Model Solution Section 14.5 14.6. 14.5. Use the Chain Rule to find dz where z cosx + 4y), x 5t 4, y 1 t. dz dx + dy y sinx + 4y)0t + 4) sinx + 4y) 1t ) 0t + 4t ) sinx + 4y) 0t + 4t ) sin 5t 4 + 4 ) t 14.5.5 Use the Chain Rule to find dw where w xe y/z, x t, y 1 t, z 1 + t. dw w dx + w dy y + w dz e y/z t + x z ey/z 1) + xy ) z e y/z t x z xy ) ) z e y/z t t 1 + t t 1 t) 1 + t) e 1 t)/1+t) t1 + t) t 1 + t) t ) 1 t) 1 + t) e 1 t)/1+t) t + 5t + 8t ) 1 + t) e 1 t)1+t) 14.5.10 Use the Chain Rule to find and s t where z e x+y, x s t, y t s. 1
MATH 004 Homework Solution s s + y y s e x+y 1 t + ex+y 1 t t ) s e s t + t s t ) 1 s t t ) s e x+y t t + y y t e x+y s ) t + e x+y 1 s t + ) e s t + t s s s s t + ) e x+y s 14.5.11 Use the Chain Rule to find and s t where z e r cos θ, r st, θ s + t. s r r s + θ θ s e r cos θ t + e r s sin θ) t s + t cos θ t cos ) s + t s ) s + t sin s + t ) s s + t sin θ ) e r e st t r r t + θ θ t e r cos θ s + e r t sin θ) s s + t cos θ s cos ) s + t t ) s + t sin s + t ) t s + t sin θ ) e r e st 14.5. Use the Chain Rule to find w r, w θ where when r, θ π. w xy + yz + zx, x r cos θ, y r sin θ, z rθ, w r w r + w y y r + w r y + z) cos θ + x + z) sin θ + y + x)θ
MATH 004 Homework Solution r, θ π x cos π 0, y sin π, z π π w + π) cos π r + 0 + π) sin π + + 0)π π w r r,θ π w θ r,θ π w θ + w y y θ + w θ y + z) r sin θ) + x + z)r cos θ + y + x)r + π) sin π ) + 0 + π) cos π + + 0) π 14.5.6 Use the Chain Rule to find α, β, γ where when α 1, β, γ 1. u xe ty, x α β, y β γ, t γ α, α α + y y α + t t α e ty αβ + xte ty 0 + xye ty γ αβ + xyγ )e ty α 1, β, γ 1 x 1), y 1 4, t 1 1) 1 α 1) + 4 1 )e 1 4 4e 4 α 1,β,γ1 β β + y y β + t t β e ty α + xte ty βγ + xye ty 0 α + xtβγ)e ty β 1) + 1) 1)e 1 4 7e 4 α 1,β,γ1 γ γ + y y γ + t t γ e ty 0 + xte ty β + xye ty γα xtβ + xyγα)e ty γ 1) + 4 1 1))e 1 4 4e 4 α 1,β,γ1 14.5.9 The length l, wih w, and height h of a box change with time. At a certain instant the dimensions are l 1 m and w h m, and l and w are increasing at a rate of m/s while h is decreasing at a rate of m/s. At that instant find the rates at which the following quantities are changing.
MATH 004 Homework Solution a) The volume: From the conditions above, dv Volume V lwh l0) 1, w0), h0), dl dw, dh,. dv V dl l + V dw w + V h wh dl +lh dw b) The surface area: dh whdl + lhdw dh + lw +lw dh +1 +1 ) 6 m /s) Surface area A lw + lh + wh da A dl l + A dw w + A dh w + h)dl + l + h)dw + l + w)dh h da w + h) dl + l + h) dw + l + w) dh + ) + 1 + ) + 1 + ) ) 10 m /s) c) The length of a diagonal: dd Diagonal D l + w + h dd D dl l + D w dw + D dh h l dl l + w + h + w dw l + w + h + h dh l + w + h l dl w dw l + w + h + h dh l + w + h + l + w + h 1 1 + + + 1 + + + ) 0 m/s) 1 + + 14.5.4 One side of a triangle is increasing at a rate of cm/s and a second side is decreasing at a rate of cm/s. If the area of the triangle remains constant, at what rate does the angle between the sides change when the first side is 0 cm long, the second side is 0 cm, and the angle is π 6? Let x be the length of the first side, y be the length of the second side, and let θ be the angle between them. then from the assumption, x0) 0, y0) 0, θ0) π 6, dx dy,. The area of the triangle is A : 1 xy sin θ. 4
MATH 004 Homework Solution Because the area is constant, da 0. By the Chain Rule, 0 da A dx + A dy y + A dθ y dx sin θ + x dy sin θ + 1 dθ xy cos θ 0 sin π 6 + 0 sin π 6 ) + 1 0 0 cos π dθ 6 5 + 150 dθ dθ 5 00 1 1 rad/s) 14.5.45 If z fx, y), where x r cos θ and y r sin θ, a) find and r θ. r r + y y r cos θ + y sin θ b) Show that θ θ + y y ) + θ ) y ) cos θ + y sin θ + 1 r ) cos θ + y + 1 ) r r sin θ ) cos θ + sin θ ) + r sin θ) + y r cos θ ) + 1 r r ). θ r sin θ) + ) y r cos θ ) sin θ cos θ sin θ + y ) y r sin θ cos θ + r cos θ) y ) cos θ sin θ cos θ sin θ)+ sin θ + cos θ ) y y ) ) + y 14.6.6 Find the directional derivative of fx, y) e x cos y at 0, 0) in the direction indicated by the angle θ π 4. Direction vector u: cos π 4, sin π 4 1 1, f f x, f y e x cos y, e x sin y 5
MATH 004 Homework Solution f0, 0) e 0 cos 0, e 0 sin 0 1, 0 D u f0, 0) f0, 0) u 1, 0 1 1, 1 14.6.8 a) Find the gradient of fx, y) y x. b) Evaluate the gradient at P 1, ). f f x, f y y x, y x f1, ) 1, 4, 4 1 c) Find the rate of change of f at P in the direction of the vector u 1 i+ 5j). D u f1, ) f1, ) u 4, 4 5, 8 + 4 5 14.6.15 Find the directional derivative of fx, y, z) xe y + ye z + ze x at the given point 0, 0, 0) in the direction of v 5, 1,. The unit vector to the direction of v: v 5 + 1 + ) 0 u v v 5 1,, 0 0 0 f f x, f y, f z e y + ze x, xe y + e z, ye z + e x f0, 0, 0) e 0 + 0e 0, 0e 0 + e 0, 0e 0 + e 0 1, 1, 1 D u f0, 0, 0) f0, 0, 0) u 1, 1, 1 5 1,, 4 0 0 0 0 14.6.4 Find the maximum rate of change of fx, y, z) x + y z direction in which it occurs. at 1, 1, 1) and the f f x, f y, f z 1 z, 1 z, x + y z f1, 1, 1) 1 1, 1 1. 1 + 1 1, 1, 1) The maximum rate of change occurs to the direction of f1, 1, 1) 1, 1,. The unit vector in this direction is 1, 1,.) In this case, the maxi- 6 6 6 mum rate of change is f1, 1, 1) 1) + 1) + ) 6. 6
MATH 004 Homework Solution 14.6.5 Let f be a function of two variables that has continuous partial derivatives and consider the points A1, ), B, ), C1, 7), and D6, 15). The directional derivative of f at A in the direction of the vector AB is and the directional derivative at A in the direction of AC is 6. Find the directional derivative of f at A in the direction of the vector AD. AB, 1,, 0 The unit vector u to the direction of AB is i. AC 1, 7 1, 0, 4 The unit vector v to the direction of AC is j. If f1, ) a, b, D u f1, ) f1, ) u a, b i a 6 D v f1, ) f1, ) u a, b j b f1, ), 6 AD 6, 15 1, 5, 1 AD 5 + 1 169 1 The unit vector w to the direction of AD is 5 1, 1 1. D w f1, ) f1, ) w, 6 5 1, 1 1 7 1 14.6.4 a) Find an equation of the tangent plane to xyz 6 at,, 1). A point on the tangent plane:,, 1) fx, y, z) xyz f f x, f y, f z yz, xz, xyz f,, 1) 1, 1, 1,, 1 A normal vector:,, 1 An equation of the tangent plane: x ) + y ) + 1z 1) 0 or x + y + 1z 4 7
MATH 004 Homework Solution b) Find equations of the normal line to xyz 6 at,, 1). A point on the normal line:,, 1) A direction vector: f,, 1),, 1 Symmetric equations of the normal line: x y z 1 1 14.6.51 Show that the equation of the tangent plane to the ellipsoid x a + y b + z c 1 at the point x 0, y 0, z 0 ) can be written as A point on the plane: x 0, y 0, z 0 ) xx 0 a + yy 0 b + zz 0 c 1. fx, y, z) x a + y b + z c f f x, f y, f z x a, y b, z c fx 0, y 0, z 0 ) x 0 a, y 0 b, z 0 c A normal vector: fx 0, y 0, z 0 ) x 0 a, y 0 b, z 0 c or x 0 a, y 0 b, z 0 c An equation of the tangent plane: x 0 a x x 0) + y 0 b y y 0) + z 0 c z z 0) 0 x 0 x a x 0 a + y 0y b y 0 b + z 0z c z 0 c 0 xx 0 a + yy 0 b + zz 0 c x 0 a + y 0 b + z 0 c Because x 0, y 0, z 0 ) is on the ellipsoid, xx 0 a + yy 0 b + zz 0 c 1. 14.6.55 Are there any points on the hyperboloid x y z 1 where the tangent plane is parallel to the plane z x + y? The tangent plane at x 0, y 0, z 0 ) is parallel to the plane z x + y when the normal vector fx 0, y 0, z 0 ) is parallel to 1, 1, 1, or equivalently, fx 0, y 0, z 0 ) c 1, 1, 1 for some constant c. fx, y, z) x y z f f x, f y, f z x, y, z fx 0, y 0, z 0 ) x 0, y 0, z 0 c 1, 1, 1 8
MATH 004 Homework Solution x 0 c, y 0 c, z 0 c x 0 c, y 0 c, z 0 c Because x 0, y 0, z 0 ) is on the hyperboloid, 1 x 0 y 0 z 0 c ) c ) c ) c 4 The right hand side is always non-positive. Therefore there is no solution and there is no such point. 14.6.59 Where does the normal line to the paraboloid z x + y at the point 1, 1, ) intersect the paraboloid a second time? fx, y, z) x + y z 0 f x, y, 1 f1, 1, ),, 1 A direction vector:,, 1 A point on the line: 1, 1, ) A parametric equation of the normal line: rt) 1, 1, + t,, 1 1 + t, 1 + t, t frt)) 1 + t) + 1 + t) t) 0 1 + 4t + 4t ) + t 0 8t + 9t 0 t 0, 9 8 Another intersection point: r 9 8 ) 5 4, 5 4, 5 8 14.6.6 Show that the pyramids cut off from the first octant by any tangent planes to the surface xyz 1 at points in the first octant must all have the same volume. fx, y, z) xyz f yz, xz, xy fx 0, y 0, z 0 ) y 0 z 0, x 0 z 0, x 0 y 0 The tangent plane at x 0, y 0, z 0 ): y 0 z 0 x x 0 ) + x 0 z 0 y y 0 ) + x 0 y 0 z z 0 ) 0 xy 0 z 0 x 0 y 0 z 0 + yx 0 z 0 x 0 y 0 z 0 + zx 0 y 0 x 0 y 0 z 0 0 xy 0 z 0 + yx 0 z 0 + zx 0 y 0 9
MATH 004 Homework Solution because x 0 y 0 z 0 1. The intersection of tangent plane and x-axis: The intersection with y-axis: y z 0 x y 0 z 0 x z 0 y x 0 z 0 The intersection with z-axis: x y 0 z x 0 y 0 The area of the base of the pyramid: 1 y 0 z 0 x 0 z 0 9 x 0 y 0 z 0 9 z 0 The height of the pyramid: The volume of the pyramid: x 0 y 0 1 area of the base height 1 9 z 0 x 0 y 0 9 x 0 y 0 z 0 9 Therefore it is independent from the choice of a point x 0, y 0, z 0 ) on the surface. 10