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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 xdirection at this point? Very brieﬂy justify
your answers. According to the contour in graph in (a), the zvalues 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 xdirection 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 xdirection (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 xdirection? What is the slope of the plane in the ydirection? 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 ydirection 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, —a2—j2:(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(1x2)e_(igﬁ),
%§(x,y)=yx(x —3)e—(i§ﬁ]
azf(x,y)=(l—y2)(1x2)e[ ...
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 Lewis

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