Homework 8 Model Solution Section

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1 MATH 004 Homework Solution Homework 8 Model Solution Section 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 ) t 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) Use the Chain Rule to find and s t where z e x+y, x s t, y t s. 1

2 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 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 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

3 MATH 004 Homework Solution r, θ π x cos π 0, y sin π, z π π w + π) cos π r 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 ) π) cos π + + 0) π 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) )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) ))e 1 4 4e 4 α 1,β,γ 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.

4 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 ) 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 + ) ) ) ) 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 ) 0 m/s) 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

5 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 sin π 6 ) cos π dθ dθ dθ rad/s) 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 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

6 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, 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, 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 ) 0 u v v 5 1,, 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,, 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) 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 mum rate of change is f1, 1, 1) 1) + 1) + ) 6. 6

7 MATH 004 Homework Solution 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 The unit vector w to the direction of AD is 5 1, 1 1. D w f1, ) f1, ) w, 6 5 1, 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

8 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 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 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

9 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 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,, t, 1 + t, t frt)) 1 + t) t) t) t + 4t ) + t 0 8t + 9t 0 t 0, 9 8 Another intersection point: r 9 8 ) 5 4, 5 4, 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

10 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