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UCSC / Math / MATH 23 / math 23a study guides

# math 23a study guides Description

##### Uploaded: 10/29/2016
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## What must you conclude ? ## b) If these conditions hold, what is the derivative Df (~v ) ? ## What must be true about L and E ? 23A16 REVIEW 2 SOLUTIONS 1. a) L : Rn → Rm is a linear transformation. The error term must satisfy lim ~h→~0 kEk k~hk= 0. b) If these conditions hold then Df (~v ) = L . L is represented by an m × n matrix whose ij th entry is ∂fi ∂xj(~v ). c) We have a theorem that says if f is a C1function then f is differentiable. To be C1 means that eveWe also discuss several other topics like bme ucsc
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ry one of the mn partial derivatives ∂fi ∂xj(~x) (before evaluation at ~x = ~v ) are themselves continuous functions. 2. a) No conclusion about the existence of limit or continuity at (0,0) is possible from this information. b) By following the curve x = y2to (0,0) you will get a limit of 1/2 which is not equal to the limiting value 0 from part a, therefore this function has no limit and hence is not continuous at (0,0). c) To compute the partial with respect to x you need to find lim f (h, 0) − f (0, 0) for the y partial are zero, in particular, they both exist. h→0 h. This limit as well as the one d) No ! Differentiability is stronger than continuity and by part b, f is not continuous at (0, 0) so f is not differ entiable at (0, 0). e) Since continuous partials imply a function is differentiable, it must be that the partials (as functions in their own right) are not continuous at (0,0). 3. The idea here is to recognize the limit as a partial derivative. Since the change vector is h~k, change is occurring in the z-direction only, so the limit is ∂f ∂z (2, 0, π) = −2. 4. As usual for a line we need a point on the line and a vector parallel to the line. The point will be where the line intersects the graph, (1, 2, −1), and the normal vector to the graph at this point, (2, 4, 1), will serve as the direction vector. The line equation is given by ~l(t) = (1, 2, −1) + t(2, 4, 1).   5. a) N~ = −∂f ∂x (1, 2), −∂f ∂y (1, 2), 1   = (−8e4, −9e4, 1). b) T(x, y ) = 2e4 + 8e4(x − 1) + 9e4(y − 2) c) The normal vector to the paraboloid will be (−2x, −2y, 1). Set this equal to the normal vector found in part a and solve for x and y . The z coordinate is found by evaluating f at this point. You should get (4e4,92e4,145 6. Set ∂f ∂x (x, y ) = 2 and ∂f 4e8). 8 3 + 2(x −43) + 4(y −23). ∂y (x, y ) = 4, solve for (x, y ) = ( 43,23), and assemble the tangent plane T(x, y ) = 7. The tangent plane equation is T(x, y ) = 8 + 16(x − 1) + 12(y − 2). Now we need two points for the line. One is (1, 3, 20) which lives on the tangent plane (check). To get one more, evaluate the tangent plane equation at (2, 1), obtaining (2, 1, 12). Now you can build the line equation. L~(t) = (1, 2, 20) + t(−1, 2, 8) works. 8. yzxyz−1 + log(x)xyzzy z−1 + log(xy)xyz   π/4 1+ π216 1 1+ π2160   c) D~r (0) =  1  9. a) Dg(0, 0) = (0 1e) b) DF (1, π/4, 0) = 0 0 π4 0 0 π4  0 −1    d) Df (1, 2, 3, 4) = 6 4 6 4 e) Df (1, 2, 3, 4) = 12 1 0 0 0 3 2 0 0 0 4 3 4 0 0 1   f) Since f is linear, f (~v + ~h) = f (~v ) + f (~h) which is of the form f (~v + ~h) = f (~v ) + L(~h) + E(~v, ~h) with f = L and E = ~0 . With no error term the condition that the error goes to zero sufficiently fast is trivially satisfied in our base definition of what it means for f to be differentiable. Since Df (~v ) = L and L = f it must be that for any ~v , Df = f . That is, a linear function is its own derivative, its derivative matrix is the same as its own matrix representation (since it’s linear).   10. a) (0 1e)  x y   =yec)   t 0 −t   d) 24+6 4 6 4   x − 1 y − 2 z − 3 = 24+6(x −1)+4(y −2)+6(z −3)+4(w −4) e)   2 6 12 4  +   2 1 0 0 0 3 2 0 0 0 4 3 4 0 0 1     x − 1 y − 2 z − 3 w − 4   w − 4 11. Any function of the form f (x, y ) = (x + 4y + a, 2x + 5y + b, 3x + 6y + c) where a, b, and c are con stants will work. 12. Rπ/4(x, y ) =  √12− √12 √12√12 !  x y   =   √x2− √y2 √x2+ √y2 ! =  x√2−y√2,x√2+y√2 . Now finding the composition f ◦ Rπ/4(x, y ) = 2( √x2− √y2)( √x2+ √y2) = x2 − y2 = g(x, y ). 14. a) ~r (t) = (2 cos t, 2 sin t) b) ~r (t) = (2 sin t, 2 cos t) c) ~r (t) = (2 cos t, 2 sin t) + (4, 7) 15. a) ~r (t) = (1 − 3t, 2 − 2t, 3 + 4t) b) ~r (t) = (t, t2) c) ~r (t) = (3 cos t, 5 sin t), t ∈ [0, 2π] 16. a) ~r (t) = (3, 4, 5) + (2 cos t, 0, 2 sin t) b) ~r (t) = (−1, 4, 7) + (0, 2 cos t, 5 sin t) c) ~r (t) = (t, t2, t2 + t4) 17. a) The point will oscillate between (-1,1) and (1,1) on the parabola y = x2. b) For 1 ≤ t < ∞ the point moves slower and slower from (0,0) along the positive side of the parabola y = x2, the entire negative branch of the parabola is covered for 0 < t < 1. c) For t = 1 the point is at the origin with speed 1. For t = e the point arrives at (1, 1) with speed √5/e. For t = e10 the point finally makes it to (10, 100) with speed √401/e10. This should give you more insight into how slow logarithmic growth is. As t decreases from 1 to e−10 the point arrives at (−10, 100) with the incredible speed e10√401. From t = e−10 to 0, the point traverses the rest of the negative branch. 18. a) Yes, ~r (t) · ~r 0(t) = 0. The speed is R|θ0(t)|. b) Not in general, ~v (t) · ~a(t) = R2θ0(t)θ00(t). c) If θ0(t) = ω, then θ00(t) = 0 and ~v (t) · ~a(t) = R2θ0(t)θ00(t) = 0 and ~a(t) = −ω2~r (t). 19. ~r (t) = ~l(t) + (0, cos(2πt), 2 sin(2πt)). 20. a) ||~r 0(t)|| =√w2 + 4t2 b) ~r (t) ·~r 0(t) = 2t3so yes, when t = 0. c) L~(t) = 1, w t, (4π/w)2) + t(8π/w)  d) Set the z-component of the line to zero, obtaining t = −2π/w. Now evaluate the line at this t, obtaining (1, −2π, 0). 21. Using ~p(t) = (cos t, sin t) show (~r (t) − ~p(t)) · ~p(t) = 0 and (~r (t) − ~p(t)) · ~r0(t) = 0. 22. T(t) = ~σ(θ) + t~σ 0(θ) = (Rθ − R sin θ + t(R − R cos θ), R − R cos θ + tR sin θ) = (x(t), y (t)). Now set x(t) = Rθ solve for t and check that at this t , y (t) = 2R. 28. (f ◦ ~c)0(0) = Df (~c(0))~c0(0) =  0 e4 8 0    0 2   =  2e4 0   29. D(f ◦ g)(1, 1) =   π 3 3π e3 0 3e3 0 −3 0 −190 0 0 0 0  2 2 0 0 1 20   =   7π 22π ,7e3 22e3 0 0 −29 −29   30. Use ∂f ∂r ∂f ∂θ  = ∂f ∂x ∂f ∂y    cos θ −r sin θ sin θ r cos θ   0 0 223. The satellites motion relative to the star is obtained by adding to the planets motion, the motion of the moon relative to the planet, and then the satellites motion relative to the moon. ~s(t) = (10 cos t, 10 sin t) + (3 cos 8t, 3 sin 8t) + (cos 4t, sin 4t). The resulting motion represented by the black curve is quite complicated. The planet, moon, and satellite vectors are shown at t = 0 and t = π/3. The red circle is the planets orbit, blue moon, and purple the satellites. 24. b) Using the hint and differentiating both sides gives 0 = ~r 0(t) · ~r (t) + ~r (t) · ~r 0(t) = 2~r 0(t) · ~r (t) so ~r (t) and ~r 0(t) must be perpendicular. c) If the particle is on the sphere then its position vector has constant length equal to the sphere’s radius so the result follows from part b. 26. a) ~r (0) = M~ (π) = (π2, 1, 10, 000). ~v (0) = M~ 0(π) = (2π, 0, 20). b) ~v (t) = (2π, 0, −32t + 20). ~r (t) = (2πt + π2, 1, −16t2 + 20t + 10, 000). c) Set 2πt + π2 = 10π2 and solve for t. Evaluate ~r (t) at this value. 27. The ellipse lifted up by 10 is parametrized by (0, 3 cos t, 10 sin t+10), the circle is parametrized by (3 cos 2t, 3 sin 2t, 0) (note the factor of 2 so the period will be π). The parametrization for the ride will be the sum of these two R~(t) = (3 cos 2t, 3 cos t + 3 sin 2t, 10+ 10 sin t). Now we need to find R~(π/2) and R~ 0(π/2), these will serve as the initial position ~r (0) and initial velocity ~v (0). (Note that it’s easier to restart the clock so that t = 0 corresponds to the instant of ejection.) R~(π/2) = (−3, 0, 20) = ~r (0), and R~ 0(π/2) = (0, −9, 0) = ~v (0). Now that we have the initial conditions we can forget about the ride and proceed as in problems 12 and 13. Integrating once gives ~v (t) = (0, −9, −32t), integrating again gives ~r (t) = (−3, −9t, −16t2 + 20). Finally, to find where impact will be set z(t) = −16t2 + 20 = 0 ⇒ t =√5/2, ~r (√5/2) = (−3, −9√5/2, 0). 323AF2016 DERIVATIVE / CURVE REVIEW PROBLEMS (Problems marked with (*) are previous exam problems.) 1*. a) Recall that the base definition for the function f : D ⊂ Rn → Rm to be differentiable at ~v ∈ D involved an expression f (~v + ~h) = f (~v ) + L(~h) + E(~v, ~h). What must be true about L and E ? b) If these conditions hold, what is the derivative Df (~v ) ? c) Give conditions (involving the partial derivatives) that imply f is differentiable at ~v . 2*. Consider the strange function  f (x, y ) =  xy 2 x2 + y4(x, y ) 6= (0, 0) 0 (x, y ) = (0, 0) A view from above and from the side are shown below. a) Find the limit of this function as (x, y ) → (0, 0) along any line y = ax. You should get 0. Does this mean f is continuous at (0, 0) ? b) Try to find another approach path to the origin that results in a non-zero limit. What must you conclude ? c) Show that ∂f ∂x and ∂f at (0, 0).) ∂y both exist at (0,0). (You must use the definition of the partials to compute their value d) Is f (x, y ) differentiable at (0, 0) ? (Think about part b.) For this function notice that just because the partials exist does not mean that ( ∂f ∂x ∂f ∂y )|(0,0) = (0 0) is the derivative. (This would mean that the tangent plane z = 0 would be a good approximation for the function around (0, 0) which is certainly not the case.) e) What must be true about the partials at (0, 0) ? f (~vo + h ~k) − f (~vo) 3*. If f (x, y, z) = x sin z + z sin y compute lim h→0 hif ~vo = (2, 0, π). 4*. Find the equation for the line that is normal to the graph of f (x, y ) = 4 − x2 − y2 where (x, y ) = (1, 2). 5*. Let f (x, y ) = yexy 2 a) Find a normal vector to the graph of f (x, y ) at (1, 2, 2e4). b) Find the equation for the tangent plane to this graph when (x,y) = (1,2). c) What point on the paraboloid z = x2 + y2 has a tangent plane parallel to the plane found in part b ? 6*. Find the equation for the tangent plane to f (x, y ) = x2 − xy + 4y2 having slope 2 in the x-direction and slope 4 in the y-direction. 7. Find the equation for the line L contained in the tangent plane to f (x, y ) = x2y3 at (1,2) if it passes through the point (1,3,20) and passes directly above (2,1). ∂x +∂f 8. If f (x, y, z) = xyzfind ∂f ∂y +∂f ∂z 19*. Compute the derivatives of the following functions at the given points. a) g(x, y ) = log(y + e) −px2y4 + 1 at (0,0). b) F (x, y, z) = (arctan(xy ), sin(y z), log(xy z + 1)) at (1,π4, 0) c) ~r (t) = (etsin t, t4, cos(π2 + t)) at t=0 d) f (x, y, z, w) = xy + y z + zw + w x at (1,2,3,4) e) f (x, y, z, w) = (xy, y z, zw, w x) at (1,2,3,4) f) If g : Rn → Rm is a linear function, what is its derivative at any point ? 10. The first order approximation to a differentiable function near the point ~xo is given by f (~x) ≈ f (~xo) + Df (~xo)(~x − ~xo) Find the first order approximation to the functions given in 20.a,c,d,e where ~xo is the given point. 11*. What is a function whose derivative at any point is   1 4 2 5 3 6   12. Consider the transformation Rθ : R2 → R2that rotates vectors counter-clockwise about the origin by angle θ. Convince yourself that Rθ(~v + ~w) = Rθ(~v ) + Rθ( ~w) and Rθ(α~v ) = αRθ(~v ) so Rθ is a linear transformation. Show that its matrix representation is Rθ =  cos θ − sin θ sin θ cos θ   . Let f (x, y ) = 2xy and g(x, y ) = x2 − y2. Show that g(x, y ) = f ◦ Rπ/4(x, y ). This shows that the graphs of these two functions look the same, that is, they both look like saddles, they coincide by rotating by π/4. 13. a) Let f : R → R be given by f (x) = x2. Describe the graph and level sets for f . b) Let f : R2 → R be given by f (x, y ) = x2 + y2. How does the graph of f intersect the (x, z)-plane and the (y, z)-plane ? Describe the graph and level curves for f . c) Let f : R3 → R be given by f (x, y, z) = x2 + y2 + z2. The graph of f = {(x, y, z, w)|(x, y, z) ∈ R3, w = f (x, y, z)} lives in the 4 dimensional (x,y,z,w)-space. How does the graph intersect (y, z, w)-space and (x, y, w)- space and (x, z, w)-space ? Describe the level surfaces for f . 14*. Consider the circle, C, centered at the origin with equation x2 + y2 = 4. a) Find a parametrization for C giving it a counter-clockwise orientation starting at (2, 0). b) Find a parametrization for C giving it a clockwise orientation starting at (0, 2). c) Find a parametrization for C if it is now centered at (4, 7). 15*. Give a parametrization for the following curves. a) The line passing through (1, 2, 3) and (−2, 0, 7). b) y = x2. c) The ellipse x29 +y2 16. Give a parametrization for the following curves. a) A circle of radius 2 with center (3, 4, 5) parallel to the (x, z)-plane. 25 = 1. b) An ellipse with a = 2 and b = 5 centered at (-1,4,7) and parallel to the (y, z)-plane. c) The curve on the graph of f (x, y ) = x2 + y2 whose (x, y )-coordinates are on the graph of f (x) = x2. 17. a) Describe the motion of a point if it’s position vector is ~r (t) = (sin t, sin2t). b) Describe the motion of a point if it’s position vector is ~r (t) = (log t, log2t), t > 0. c) What are the position and speed of the point (from part b) when t = 1, e, e10, e−10? 18. Consider a particle with position vector ~r (t) = (R cos θ(t), R sin θ(t)). a) Is the position vector of the particle perpendicular to it’s velocity vector? What is the particle’s speed? b) Is the velocity vector perpendicular to the acceleration vector? c) If the motion is uniform (θ0(t) = ω(constant)), show that ~a(t) ⊥ ~v and ~a(t) = −ω2~r (t). 19. Compound motion can be described using vector addition. Find a tracking vector (position vector) for a small object traveling counter-clockwise (c-c-w) on an ellipse parallel to the y, z-plane with a = 1 and b = 2 taking 1 time unit to complete an orbit where the center of the ellipse is moving on a line, the center’s position is given by ~l(t) = (1, 2, 3) + t(4, 5, 6). At t = 0 the object’s position is (1, 3, 3). 220*. The position vector for a particle moving on a helix is given by ~r (t) = (cosw t, sinw t, t2). a) Find the speed of the particle at time t. b) Is ~r (t) ever orthogonal to ~r 0(t) ? c) Find an equation for the tangent line to ~r (t) at t =4πw. d) Where will this line intersect the x, y plane ? 21*. Consider the spiral given by ~r (t) = (cos t + t sin t, sin t −t cos t). Let ~p(t) describe a point moving counter-clockwise around a unit circle with unit speed that is centered at the origin and ~p(0) = (1, 0). Show that for any t, the line connecting ~p(t) to ~r (t) is tan gent to the circle at ~p(t) and perpendicular to the spiral at ~r (t). The diagram should make this clear. 22. Recall the cycloid, the curve generated by a fixed point on a rolling circle of radius R. At θ = 0 the point is at (0, 0) and after rotation by θ we found the point to be at ~σ(θ) = (Rθ − R sin θ, R − R cos θ). Note that the cycloid is really another example of compound motion. One motion is clockwise rotation on a circle of radius R starting at the bottom, ~r1(θ) = (−R sin θ, −R cos θ) (check this). The other motion is the center of the circle moving on the line ~r2(θ) = (Rθ, R) . Then ~σ(θ) = ~r1(θ) + ~r2(θ). Show that the tangent line to the cycloid for any θ passes through the top of the rolling circle. Hint: You need to find the coordinates of the top of the rolling circle after rotation by θ, you should get (Rθ, 2R). Now show that your tangent line passes through this point. 23. Suppose that a planet P moves counter-clockwise in the x, y -plane around a star S in a circle of radius 10 units, completing one revolution in 2π units of time. Suppose that a moon M moves counter-clockwise around planet P in a circle of radius 3 units, completing one revolution in π/4 time units and a satellite s moves counter-clockwise in a circle of radius 1 around the moon completing one revolution in π/2 units of time. The star S is fixed at the origin and at t = 0 the planet P is at the point (10, 0) , the moon M is at the point (13, 0), and the satellite is at the point (14, 0). Find a parametric curve representing the position of the satellite s relative to the star S. (Think vector addition. All motion is in the x, y -plane.) 24. Let ~s(t), ~r (t) be differentiable vector valued functions tracing curves in R3. a) Show that the product rule holds for differentiating the product ~s(t) · ~r (t). b) Show that if ~r (t) has constant length then ~r (t) · ~r 0(t) = 0. Hint: ||~r||2 = ~r (t) · ~r (t) = c. 3c) Let ~r (t) be the position vector of a particle constrained to move on a sphere centered at the origin. Show that the particle’s velocity vector is perpendicular to ~r (t). 25. Suppose we want to ’recover’ the curve whose tangent we know. Say ~v (t) = (x(t), y (t), z(t)) and we want the vector ~r whose tangent is ~v . Then ~r (t) =  Z x(t)d t, Z y (t)d t, Z z(t)d t  Each indefinite integral generates a constant. These constants can be determined if we know ~r (0) (initial position). Now suppose we know that ~a(t) = −32~k and ~r (0) = (xo, yo, zo), ~v (0) = (v1, v2, v3) the initial position and the initial velocity. By integrating twice show that the equation of motion is ~r (t) = (xo + tv1, yo + tv2, −16t2 + v3t + zo). So if you know the acceleration and initial position and velocity you can recover the position vector (equation of motion). 26*. If there was one thing Andrea feared more than heights, it was motorcycles. So it was quite a surprise when she accepted my offer to ride on the back of my Moto Guzzi on a high speed trip across the Sierras. On one particularly treacherous section of road our path was given by M~ (t) = (t2, cos(4t), sin(20t) + 10, 000). Poor Andrea chose the wrong time (t = π) to adjust her sunglasses and was ejected tangential to our path. She was subject to gravitational forces (~a(t) = −32~k) thereafter. a) Find Andrea’s position and velocity vectors the instant she is ejected. b) Using the vectors found in part a as her initial position ~r (0) and initial velocity ~v (0), calculate her position vector ~r (t) after ejection. c) Luckily for Andrea, she was wearing a Velcro body suit. An immense plane of Velcro, having equation x = 10π2 happened to be in place. Where will Andrea stick on the plane ? 27*. Consider the amusement ride consisting of two motions. There is counter-clockwise motion on the ellipse y29+z2 100 = 1 that has been shifted upward to have center at (0, 0, 10). It takes 2π seconds to complete a revolution. Attached to this el lipse is a circle of radius 3 in a plane parallel to the x, y -plane, going counter-clockwise, taking π seconds to complete a rev olution. At t = 0 you enter the ride on this circle at loca tion (3,3,10). At t = π/2 you are ejected tangentially and subject to gravitational acceleration ~a(t) = (0, 0, −32). Where should the ground based rescue team place an air mattress ? Assume the ground is the x, y -plane, z-axis up. Picture to the right. 28*. Define f : R2 → R2 by f (x, y ) = (yex2+y, 2x2 − y2). Let ~c : R → R2 be a path given by ~c(t) = (2 cos t, sin2t + 2t). Find (f ◦ ~c)0(0). 29. Let f : R3 → R5 be given by f (x, y, z) = (xy z, exz , x sin y, 1x, 17) and g : R2 → R3is given by g(x, y ) = (y2 + 2x, π, √x). a) Compute Dg(1, 1). b) Compute Df (3, π, 1). c) Compute D(f ◦ g)(1, 1). 30*. A function f (x, y ) can be expressed in polar coordinates using the composition f (r, θ) = f ◦ g(r, θ) where g(r, θ) = (r cos θ, r sin θ). Find expressions for ∂f ∂r and ∂f ∂θ in terms of ∂f ∂x and ∂f  ∂f ∂r  2 +1r2 ∂f ∂θ  2 = 4 ∂f ∂x  2 +  ∂f ∂y  2 ∂y and use these to show

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