normal and tangent acceleration

8/11/2019 Normal and Tangent Acceleration

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Your Name Your Signature

Student ID #

Math 126 Midterm 2

1 102 93 104 125 9

Total 50

You may use a scientific (not graphing) calculator. Give exact answers. No books allowed. You may use one 8 12 11 sheet of notes. Do not share notes. In order to receive credit, you must show your work and explain your reasoning. Place a box around YOUR FINAL ANSWER to each question. If you need more room, use the backs of the pages and indicate to the grader where to find

your work.

Raise your hand if you have a question or need more paper.

Dont open the test until everyone has a copy and the start of the test isannounced.

GOOD LUCK!

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1. (10 points)

(a) Find the tangential and normal components of the acceleration vector of a particlewith position function

r(t) =ti+ 2tj+t2k.

The tangential component is given by aT = r r|r| and the normal component is

given by either aN=|r r|

|r| or byaN=|a|2 |aT|2.

r(t) = 1, 2, 2t|r(t)| =

1 + 4 + 4t2 =

5 + 4t2

r(t) = 0, 0, 2aT =

0 + 0 + 4t5 + 4t2

aN=

4 (4t)25 + 4t2

or = 15 + 4t2

|4, 2, 0| =

205 + 4t2

.

(b) A particle starts at the origin with initial velocity i j+ 3k. Its acceleration isa(t) = 6ti+ 12t2j 6tk. Find its position function.

Integrate to get v(t) =3t2 +a, 4t3 +b, 3t2 +c. Since v(0) =1, 1, 3, we havea= 1, b= 1, and c= 3. That is,

v(t) = 3t2 + 1, 4t3 1, 3t2 + 1.Integrate again to getr(t) = t3 +t+e, t4 t+f, t3 + 3t+g. Sincer(0) = 0, 0, 0,we have

r(t) = t3 +t, t4 t, t3 + 3t.

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2. (9 points) Find and classify the critical points of the function

f(x, y) =x3 3x+ 2y3 9y2 + 12y+ 1The critical points are where the first derivatives are zero.

fx(x, y) = 3x2 3 = 3(x 1)(x+ 1)fy(x, y) = 6y

2 18y+ 12 = 6(y 2)(y 1).We have then four critical points: (1, 1), (1, 2), (1, 1), and (1, 2). To classify them,we must calculate D at each.

fxx(x, y) = 6x

fxy(x, y) = 0

fyy(x, y) = 12y 18D(x, y) =fxxfyy (fxy)2

= (6x)(12y 18).Note that one does NOT need a calculator to classify any of these critical points, sincewe only care about the sign ofD.

( x , y) 6x 12y 18 D classification(1, 1 ) + local max(1, 2 ) + saddle point( 1 , 1 ) + saddle point( 1 , 2 ) + + + local min

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3. (10 points) Let f(x, y) = (x2 +y)ey

21

(a) Find the tangent plane at the point (1, 2).The tangent plane at (1, 2) is given by

z

f(1, 2) =fx(1, 2)(x

1) +fy(y

2).

We have

f(1, 2) = (1 + 2)e0 = 3

fx(x, y) = 2xey

21

fx(1, 2) = 2(1)e0 = 2

fy(x, y) =

y(x2 +y)

ey

21 + (x2 +y)

yey

21

= (1)ey

21 + (x2 +y)

1

2ey

21

fy(1, 2) =e0 + (3)1

2e0 = 1 +3

2=5

2.

The plane is

z 3 = 2(x 1) +52

(y 2)or

4x+ 5y 2z= 8if you want to write it in a more standard form.

(b) Use a linear approximation to approximate f(1.05, 1.9).Simply plug in (1.05, 1.9) into the above plane equation and solve for z.

f(1.05, 1.9) z= 3 + 2(1.05 1) +52

(1.9 2) = 2.85 .

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4. (12 points)

(a) Evaluate by reversing the order of integration:10

1x

cos(y2)dydxDrawing the region, one gets the triangle with coordinates (0, 0), (0, 1), and (1, 1).

10

y0 cos y

2

dx dy= 10 x cos y

2

|y

x=0dxdy

=

10

ycos y2 dy=

10

cos udu

2

=1

2sin(1).

(b) Evaluate by reversing the order of integration:10

1y

yex2

x3 dxdy

The region is below the parabola y = x2 bounded by x = 0 and x = 1. (Recallthat x=

y is only the right half of the parabola.)

1

0

x2

0

y

x3ex

2

dy dx=

1

0

1

2

(x2)2 (0)2x3

ex2

dx

=

10

1

2xex

2

dx

=

10

1

2eu

du

2

=1

4(e1 e0) =1

4(e 1).

(c) Compute the average value ofy+ 2xey over [0, 2]

[1, 2]

The average value is

1

A([0, 2] [1, 2])

[0,2][1,2]y+ 2xeydA=

1

2

21

20

y+ 2xeydxdy

=1

2

21

2y+ 4eydy

=1

2(4 + 4e2 1 4e)

=3

2+ 2e2 2e

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5. (9 points) Find the volume bounded by the surfaces z+ 2x2 + y2 = 1, z= 2x2 y2 andthe planes y= 1, y= 1.First, we need to find the bounds of the region of integration. The planes give1 y 1. The other bounds are given by finding the intersection of the surfaces.

1 2x2 y2 = 2x2 y21 = 4x2

x= 12

So the region is [ 12 , 12 ] [1, 1]. In this region, the first surface is above the second:at (0, 0), z= 1 in the first, z= 0 in the second. Then the integral is

[ 1

2, 12][1,1]

(1 2x2 y2) (2x2 y2)dA= 11

12 1

2

1 4x2dA

=

11

dy

12

12

1 4x2dx

= 2(x 4x2|1

2

12

)

=4

3

Alternatively, you can find the integral of 1 2x2 y2 and 2x2 y2 over the regionseparately (they are 1 and

13 respectively) and subtract the second from the first.

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normal and tangent acceleration

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