Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
The exam contains 10 problems. • The last page is the formula sheet, which you may detach. • Good luck! ... by the Fundamental Theorem of Calculus.
Typology: Study notes
1 / 12
This page cannot be seen from the preview
Don't miss anything!
Name:
Circle one: Section 6 Section 7.
Question: 1 2 3 4 5 6 7 8 9 10 Total Points: 10 10 10 10 10 10 10 10 10 10 100 Score:
(a) The curve traced by 〈cos^2 (t), sin^2 (t)〉 is a circle. True False
Solution: False.
(b) The plane 3x + 2y − z = 0 is perpendicular to the line x = 3t, y = 2t, z = −t. True False
Solution: True.
(c) The function f (x, y, z) =
{sin(x+y+z) x+y+z if^ x^ +^ y^ +^ z^6 = 0 1 if x + y + z = 0 is continuous at (0, 0 , 0). True False
Solution: True.
(d) If the acceleration is constant, then the trajectory must be a straight line. True False
Solution: False.
(e) The complex number e2+3i^ has magnitude 2. True False
Solution: False.
(a) (3 points) Mark any three critical points of f. Label them A, B, and C. Identify whether they are local minima, local maxima or saddle points.
Solution: A: Saddle; B, D: Local maxima, C, E: Local minima.
(b) (2 points) Draw a vector at (1, 1) indicating the direction of ∇f at (1, 1). (c) (3 points) Determine the sign of
Solution: ∂f /∂x(1, 1) ≈ 0. 2 / 0 .2 = 1. 0
Solution: Let r(t) = 〈t^2 , ln t, t^3 〉. Then r(1) = 〈 1 , 0 , 1 〉.
r′(t) = 〈 2 t, 1 /t, 3 t^2 〉 r′(1) = 〈 2 , 1 , 3 〉.
The tangent line is the line through 〈 1 , 0 , 1 〉 in the direction of 〈 2 , 1 , 3 〉, that is:
(x, y, z) = 〈 1 , 0 , 1 〉 + t〈 2 , 1 , 3 〉.
In other words, x = 1 + 2t, y = t, z = 1 + 3t.
(b) (5 points) Write an equation of the normal plane to the curve at the same point.
Solution: The normal plane is the plane through 〈 1 , 0 , 1 〉 perpendicular to r′(1) = 〈 2 , 1 , 3 〉. The equation is
〈x − 1 , y, z − 1 〉 · 〈 2 , 1 , 3 〉 = 0 =⇒ 2(x − 1) + y + 3(z − 1) = 0 =⇒ 2 x + y + 3z = 5.
Solution: We have
Acceleration −→a = i − 10 j Therefore, velocity −→v (t) =
−→a dt
= ti − 10 tj + −→c.
Since −→v (0) = i + j, we get −→c = i + j. Therefore −→v (t) = (1 + t)i + (1 − 10 t)j
Therefore, position −→r (t) =
−→v (t) dt
t + t
2 2
i + (t − 5 t^2 )j + −→c.
Since −→r (0) = 4j, we get −→c = 4j. Therefore,
−→r (t) =
t + t
2 2
i + (4 + t − 5 t^2 )j.
In other words, at time t, the ball is at x = t + t^2 /2 and y = 4 + t − 5 t^2.
(b) (3 points) Where is the ball when it hits the ground?
Solution: The ball hits the ground when y = 0. That is
4 + t − 5 t^2 = 0 5 t^2 − t − 4 = (5t + 4)(t − 1) = 0.
The only positive solution is t = 1. So the ball hits the ground at t = 1. At this point, it is at x = 1 + 1/2 = 3/2 and y = 0.
(a) (2 points) Find the value of u when s = 2 and t = 2.
Solution: Then x = 8 and y = 0, so that exy^ = e^0 = 1.
(b) (8 points) Find an approximate numerical value of u when s = 2.01 and t = 1.98.
Solution: We know that
∆u ≈ ∂u∂s ∆s + ∂u∂t ∆t.
By the chain rule, ∂u ∂s =^
∂u ∂x ·^
∂x ∂s +^
∂u ∂y ·^
∂y ∂s = yexy(t + 1) + xexy(t − 1) = 8 at s = 2, t = 2.
Likewise, ∂u ∂t
= ∂u ∂x
· ∂x ∂t
· ∂y ∂t = yexy(s + 1) + xexy(s − 1) = 8 at s = 2, t = 2.
Hence
∆u ≈ 8 · 0 .01 + 8 · (− 0 .02) = − 0. 08.
In other words, u ≈ 1 − 0 .08 = 0.92. (By the way, the exact answer is u = 0. 92192448707... .)
Race track
Solution: Let r(t) be the position of the car at time t. We know that at time t, the tangent and normal components of accelaration are given by
aT = |r′(t)|′^ and aN = κ|r′(t)|^2 ,
where κ is the curvature of the trajectory at the point r(t). Since the speed |r′(t)| is constant, we immediately get that
aT = 0.
What remains is aN = κ|r′(t)|^2 = 100^2 κ. To get the answer, we must compute the curvature of the ellipse at the point P. To compute the curvature, we first conveniently parametrize the ellipse. For example,
x = 2 cos t, y = sin t.
Then
κ = |〈−2 sin^ t,^ cos^ t〉 × 〈−2 cos^ t,^ −^ sin^ t〉| |〈−2 sin t, cos t〉|^3 = |(2 sin
(^2) t + 2 cos (^2) t)k| (4 sin^2 t + cos^2 t)^3 /^2 = 2 (4 sin^2 t + cos^2 t)^3 /^2 = 2 at P (set t = 0).
Therefore, aN = 2 · 1002 = 20000 km/h^2.
Solution: Let the radius of the cup be r and the height h. We want
Volume = πr^2 h = 100π, that is r^2 h = 100.
We want to minimize
Surface area = πr^2 (bottom) + 2πrh(side) = π(r^2 + 2rh).
So, we want to minimize f (r, h) = r^2 + 2rh subject to g(r, h) = r^2 h = 100. By the method of Lagrange multipliers, the extremal values are attained when
∇f = λ∇g.
We have ∇f = 〈 2 r + 2h, 2 r〉, ∇g = 〈 2 rh, r^2 〉. These two are multiples of each other when 2 r + 2h 2 rh =
2 r r^2. Simplifying and cross-multiplying gives
r + h = 2h =⇒ r = h.
So the only critical point is when r = h. To find this value, we use the constraint
r^2 h = 100.
This gives r = h = 100^1 /^3. (By the way, r = h would result in a very bizarre coffee cup!)
