midterm1solutions

midterm1solutions - Name: ..._SO_L_.UTIQA_KLY____ lD #:...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Name: ..._SO_L_.UTIQA_KLY____ lD #: Section: (circle one) 4pm 5pm 6pm 7pm 8pm TA's name: Signature: Midterm Exam #1 — MAT 17C Spring 2009 Instructor: Tim Lewis Wednesday, April 22, 2009 When you receive your exam, fill in the top of this page but do not begin the exam until you are told to do so. Read each question carefully and do all your work on the exam; you must show all your work to get full credit. Use the back of each page if you need extra space. No books or notes are permitted during the exam. Cellphones must be turned off and put away. Note that the last page of the exam is a table of very helpful derivatives. 1 /25 2. /25 3. /25 4 /25 Total: /100 PROBLEM I: [25 points] The receptive field of a sensory neuron is a region of space in which the presence of a stimulus will alter the level of activity of that neuron. Suppose that the firing rate of a somatosensoryl neuron for stimuli within its receptive field is given by the function 2 V=f(x,y)=9-[x2+(%] where Vis the ‘firing’ rate of the neuron in Hz for a stimulus at particular location (x, y) on the subject’s finger for v 2 0. (a) [10 pts] On a single graph, sketch the contours of f (x, y) for v = 5 , and v =8. Be sure to label your graph. 2 Contours: V=f(x,y)=9-[x2 (b) [5 pts] Plot the point (x = -1, y = l) on your graph in (a). Based on your contour plot, is f (x, y) increasing or decreasing in the x-direction at this point? Very briefly justify your answers. According to the contour in graph in (a), the z-values of the contours are increasing in the x—direction, i .e. v = f (x, y) is increasing in the x—direction at the point (-l,l). l The somarosensory system is a diverse sensory system comprising the receptors and processing centres to produce the sensory modalities such as touch, temperature. proprioception (body position), and nociception (pain). (c) [10 pts] Calculate the partial derivative gl— and?- , and evaluate them at (x, y) = (—1,1) . x 3’ Based on your partial derivatives, is f(x, y) increasing or decreasing in the x-direction at this point? Briefly justify your answers. [How do your answers in (c) compare to your answers in (b)?] a-_f__ ii- _ a 2x ax(1,l)—2>0 8f y y 8f 1 _=_2_:__ _ _1, =_._ By 4 2 ay( 1) 2<0 The signs of the partial derivatives indicate that, at the point (-1,1), V = f (x, y) is increasing in the x-direction (and decreasing in the ydirection). Note that these conclusions in (c) agree with those in (b). PROBLEM 2: [25 points] (a) [10 pts] What is the general equation (formula) for the tangent plane of the function f (x, y) at (x0, yo) ? What is the value of the tangent plane function at (x0, yo) ? What is the slope of the plane in the x-direction? What is the slope of the plane in the y-direction? 3' 3’ i‘(~fi)’)b)(X—xo)+'—f(xo’yO)()"y0) f(x,y)-— f(xo,y0)+ a y The value of the tangent plane function at (x0, yo) is z = f(x0, yo) . The slope of the tangent plane in the x—direction at (x0, yo) is Sim), yo). x The slope of the tangent plane in the y-direction at (x0, yo) is 31mph). 3’ Suppose that z=f(x,y)=x2y+2xey. (b) [10 pts] Find the tangent plane of z = f (x, y) at (2,0). %6—=2xy+2ey %f-(2,0)=2 a; a; f(2,0)=4 ——=x2+2xe" —(2,0)=8 3y 3y z=4+2(x—2)+8(y——0) =2x+8y (c) [5 pts] Using the tangent plane, approximate z = f (x. y) at (2.1,—0.l). (Note that the actual function evaluation z = f (2.1,~0.1) = 3.359 ). z =f(2.1,-0.l)=: 2(2.l)+8(~0.1) =3.4 PROBLEM 3: [25 points] A malphogen is a substance governing the pattern of tissue development. A morphogen usually spreads from a localized source and forms a concentration gradient across a developing tissue, triggering cellular differentiation in distinct spatial order. Suppose that the concentration of a morphogen in a portion of an embryo was given by . 18 Z: xaV 7‘“ j( ') xz+4y2+l where z = f (x, y) is the concentration of a morphogen at position (x, y) in the embryo. (a) [5 pts] Determine the gradient vector W (x, y) and evaluate it at the position (—1,l). Vf(x,y)=[§£ i]: -__;3_6x__, -(36)(4y) (x2 +4y2 +1)2 (x2 +4y2 +1)2 —36<--1) -(36)(4) V (—1,1)= , f [(1+4+1)2 (1+4+1)2 law) (b) [8 pts] Find the directional derivative of f (x, y) at the position (—1,1) in the direction of W: (1,—1). Determine whether the morphogen concentration is increasing or decreasing in this direction. D“ f(—1,1) = Vf(—l,l)-(71_2—,— ) 2(1’4).( 12 7713') el— * note that i7=——=—1-(1,—1) if) llvT/ll J5 || 5 ——>0 J5 :> v is increasing in the direction of W. (c) [5 pts] Determine the direction that has the greatest rate of change in the morphogen concentration at the position (—1,!) , i.e. determine the direction that maximizes the directional derivative of f (x, y) at (—1.1). Du f (—1, 1) is maximum in the direction of the gradient vector Vf(—1,l) =(1,~4), __ Vf(——1,1) __ l r7170»). ie . . u —"Vf(_1,1) (d) [7 pts] What is the rate of change of the morphogen concentration along the path (x(t),y(t)) =(2cos(t),sin ? What does this path correspond to? Note that along (x(t),y(t)) = (2cos(t),sin Z(t)=f(x(’)ay(t)) _ 18 _ x(t)2 +4y(t)2 +1 _ 18 _ 18 _ _ 4cos2(t)+4sin2(t)+1_ 4+4+1— That is, 2(t) is constant along the path (x(t),y(t)) = (2cos(t),sin(t)) , and therefore the rate of change of the morphogen concentration is zero. This implies that the path (x(t),y(t)) =(2cos(t),sin(t)) is a contour of the function f(x,y) (with z=2). PROBLEM 4: [25 points] Suppose that the response of subjects to a combination of two drugs for hypertension is described by the function 1 V U41") 7 z=f(x.y)=xye-[ ' , where z is the change in a subject’s blood pressure in response to x mg of drug X and y mg of drug Y. (a) [15 pts] Determine all of the “physical” critical points of the function z = f (x, y) (i.e. the critical points with x 2 0 and y 2 0 ), and classify them as local maxima, local minima or saddles. y(1—x2)e_(L¥_)=0 :> y=0 or x=iz1 x(l—y2)e—(L§L]=O :> x=0 or y=il Therefore, (0,0) and (1,1) are the potential critical points with x 2 O and y 2 0. 2 _ ii 2 ' _ .fli %:j2:(1,1)= yx(x2 —3)e [ l ] =—2e'1, i.:—y{—(l,1)=yx(y2 —3)e[ 2 J =-—2e*l (x.y)=(l,l) (x.y)=(l.l) 82 1 V a _ I2?.2 axaj;(1,1)=(1~y-)(1—x~)e( ) =0 (x,_\‘)=(l,l) _ 32f 821; 82f 2 _ D_ 8x2 Byz —[E)x8y _4e (.x._\')=( 1.1) D >0 and %%<0 at (1.1), therefore (1,1) is a local maximum. x iEL) =0, —a-2—j-2-:(0,0)=yx(y2—3)e-[é—ELJ =0 3y (x..\')=(0‘0) (x.y)=(0,0) 32f (0 0)=(l—y2)(1—x2)e_(i}fi] =1 Bxay , (xyy)=(0.0) _ 32f it _ 32f 2 _ _ D _ 3x2 33:2 [8x837] _ l (x.y)=(0v0) D < 0 at (0,0), therefore (0,0) is a saddle. (b) [10 pts] Suppose that the domain of z: = f (x, y) is restricted to D={(x,y):x§20, y20,)c2 +y2 S4}. Find the global maximum of f (x, y) (i.e. find the maximal response to the drugs). Note that z = f (x, y) =0 on the two straight edges of the domain x=0 and y=0, so you only have to consider the curved portion of the boundary x2 + y2 = 4. The curved portion of the boundary x2 + y2 = 22 is an arc on a circle of radius 2 and can It be described by the parameterized curve (x(t),y(t)) = (2cos(t),2sin(t)), OStS—2-. Note that along (x(t), y(t)) = (2005(t),28in(t)), 2(1) = f (x(:).y(r)) _[ X1122+VU)2 J 2 = W) N) e _(4c053(/)+4sm3m] . : 4cos(t)sin(t) e ' = 4e”2 cos(t) sin(t) Look for max/min at points where dz d — - “ ' $25,446 2C05(t)sm(t))=4e 2(cosz(t)—sm2(t))=0. Therefore, there are potential global max/min on boundary when cos2(t)—sin2(t)=0,forOStsg :> t=%, which corresponds to the point i”) VD (‘ I) x —- ,v — = —-—,—— . l 4 ‘ 4 JEJE Conclusion: (i) In the interior of the domain, there is a. local maximum of z = f(l,l) = e“1 =0.3679, (ii) on the two straight edges of the domain, z = f(x, y) = 0, and (iii) on the curve portion of the portion of the domain’s boundary, there is a maximum of z = [L L) = 2e‘2 = 0.3033. «ER/5 Therefore, the global maximum of z = f(x, 3) is 0.3679, which occurs at (1,1). Table of Derivatives: [#3] z=f(xyy)=x2+::2+1 grew”: (x2 3:102 [#4] z=f(x,y)=xy %€(x.y)=y(1-x2)e_(igfi), %§(x,y)=yx(x —3)e—(i§fi] azf(x,y)=(l—y2)(1-x2)e-[ ...
View Full Document

Page1 / 10

midterm1solutions - Name: ..._SO_L_.UTIQA_KLY____ lD #:...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online