Practice Final Exam Solutions. Calculus 3, Section 001 .... 80t2 + 100. 〈−16t2 −
20,0,8t2 + 10〉. And at t = 1, we get. B(1) = 1. 3. √. 180. 〈−36,0,18〉 = 1. √. 5.
Practice Final Exam Solutions Calculus 3, Section 001
Name: Answer the questions in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page. No calculators are permitted. Good luck, have fun! Question Points Score 1
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1. (a) (4 points) Let ~v and w ~ be two vectors in Find the angle between ~v and w. ~ Solution: Since cos θ = we see that cos θ =
R3
with |w| ~ = 4 and ~v · w ~ = 2|~v |.
~v · w ~ |~v ||w| ~
2 1 2|~v | = = |~v ||w| ~ |w| ~ 2
and so θ = π3 . (b) (4 points) Show that, for any two vectors ~a and ~b, we have (~a +~b)×(~a −~b) = 2(~b × ~a). Solution: Since the cross product has a distributive law, we get (~a + ~b) × (~a − ~b) = ~a × ~a + ~b × ~a − ~a × ~b − ~b × ~b = ~b × ~a − ~a × ~b = ~b × ~a + ~b × ~a = 2 ~b × ~a
2. (10 points) Let ~r(t) = ht+1, t2 +2, 2ti. Find the equation of the osculating plane at time t = 1. Solution: We first find r~0 (t) = h1, 2t, 2i so |r~0 (t)| =
√ 5 + 4t2
giving T (t) = So
T (t) = − 0
1 2t 2 √ ,√ ,√ 5 + 4t2 5 + 4t2 5 + 4t2
4t 10 8t , ,− 2 3/2 2 3/2 (5 + 4t ) (5 + 4t ) (5 + 4t2 )3/2
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and s
16t2 100 64t2 + + (5 + 4t2 )3 (5 + 4t2 )3 (5 + 4t2 )3
s
80t2 + 100 (5 + 4t2 )3
|T 0 (t)| = =
√ 80t2 + 100 = (5 + 4t2 )3/2
which means (5 + 4t2 )3/2 T 0 (t) 4t 10 8t =√ N (t) = 0 − , ,− |T (t)| (5 + 4t2 )3/2 (5 + 4t2 )3/2 (5 + 4t2 )3/2 80t2 + 100 4t 10 8t = −√ ,√ , −√ 80t2 + 100 80t2 + 100 80t2 + 100
So we can now compute B(t) = T (t) × N (t) 1 2t 2 = √ ,√ ,√ 5 + 4t2 5 + 4t2 5 + 4t2 4t 10 8t × −√ ,√ , −√ 80t2 + 100 80t2 + 100 80t2 + 100 1 √ (h1, 2t, 2i × h−4t, 10, −8ti) =√ 5 + 4t2 80t2 + 100 1 √ =√ (h1, 2t, 2i × h−4t, 10, −8ti) 5 + 4t2 80t2 + 100 ~i ~k ~j 1 √ =√ 1 2t 2 2 2 5 + 4t 80t + 100 −4t 10 −8t ! 2t 2 1 1 2t 2 1 ~ ~ ~ √ =√ i − j + k 2 2 5 + 4t 80t + 100 10 −8t −4t −8t −4t 10
1 √ =√ −16t2 − 20, 0, 8t2 + 10 5 + 4t2 80t2 + 100 And at t = 1, we get 1 1 B(1) = √ h−36, 0, 18i = √ h−2, 0, 1i = 3 180 5
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−2 1 √ , 0, √ 5 5
and since ~r(1) = h2, 3, 4i the equation of our plane is 1 −2 √ (x − 2) + √ (z − 4) = 0 5 5 (And in case you’re wondering, something this long will absolutely not appear on the final!) 3. (10 points) Find the equation of the osculating circle to the ellipse 4x2 +y 2 = 16 at the point (2, 0). Solution: We first have to compute the curvature at this point, for which we need a parameterization; the standard one is ~r(t) = h2 cos θ, 4 sin θi Then r~0 (t) = h−2 sin θ, 4 cos θi r~00 (t) = h−2 cos θ, −4 sin θi |r~0 (t) × r~00 (t)| = 8 sin2 θ + 8 cos2 θ = 8 p |r~0 (t)| = 4 sin2 θ + 16 cos2 θ So now |r~0 (t) × r~00 (t)| |r~0 (t)|3 8 = 2 (4 sin θ + 16 cos2 θ)3/2
κ(t) =
Now to get the point (2, 0), we can take t = 0. Then κ(0) =
8 1 = 3/2 (16) 8
1 So the osculating circle has radius which is κ(π) = 8. By the symmetry of the ellipse, we see that the circle will have its center along the x-axis. Since
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it passes through the point (2, 0) and has radius 8, the center must be at (−6, 0). This gives us the equation (x + 6)2 + y 2 = 64 4. Evaluate the following limits, or explain why they do not exist. (a) (5 points)
x3 y (x,y)→(0,0) x4 + y 2 lim
Solution: We compute the limit when restricted to the line y = mx, which admits the parameterization ht, mti. The limit becomes t3 · mt mt2 = lim =0 t→0 t4 + (mt)2 t→0 t2 + m2
lim
So our first check of whether the limit exists tells us that it’s possible. Even though the degrees of the numerator and denominator match, it turns out that this limit exists. We can see this from a Squeeze Theorem argument. uv 1 To start, let’s recall the inequality 2 ≤ (remember this came out 2 u +v 2 of manipulating the obvious inequality (u − v)2 ≥ 0). If we set u = x2 1 x2 y ≤ . Let’s first assume that and v = y, we will get the inequality 4 2 x +y 2 x, y ≥ 0. Then we get x2 y 1 ≤ 4 2 x +y 2 3 x xy ≤ 0≤ 4 2 x +y 2 0≤
x3 y = 0. (x,y)→(0,0) x4 + y 2 We need to check that this is also the case for the other possible sign combinations for x and y, but we will get similar inequalities in the other cases.
If we take the limit as (x, y) → (0, 0), we get
lim
(x2 + y 2 ) ln(1 + y 2 ) (b) (5 points) lim (x,y)→(0,0) x2 + 2y 2 Solution: Since the numerator looks to go to zero faster than the denominator, let’s try to show that the limit exists. Firstly, since y 2 ≥ 0, we can add
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x2 + y 2 to each side to get x2 + 2y 2 ≥ x2 + y 2
x2 + y 2 ≤1 x2 + 2y 2
so
Now since everything involved is positive, we can also see that the result is positive, giving x2 + y 2 0≤ 2 ≤1 x + 2y 2 We now multiply by ln(1 + y 2 ) to get 0≤
(x2 + y 2 ) ln(1 + y 2 ) ≤ ln(1 + y 2 ) x2 + 2y 2
We can now take limits as (x, y) → (0, 0), and the Squeeze Theorem will give (x2 + y 2 ) ln(1 + y 2 ) =0 lim (x,y)→(0,0) x2 + 2y 2
5. (10 points) Find the equation of the tangent plane to the function f (x, y) = x2 y − xy 2 at the point (2, 2). Solution: We have
so that
∂f = 2xy − y 2 ∂x ∂f (2, 2) = 4 ∂x
∂f = x2 − 2xy ∂y
and
∂f (2, 2) = −4 ∂y
and
Then the equation of the tangent plane is z − z0 =
∂f ∂f (x0 , y0 )(x − x0 ) + (x0 , y0 )(y − y0 ) ∂x ∂y
z − 0 = 4(x − 2) − 4(y − 2)
or
4x − 4y − z = 0
6. (8 points) Recall that the wave equation is the partial differential equation 2 ∂ 2u 2∂ u = c ∂t2 ∂x2
Show that the function u(x, t) = cos(x − ct) satisfies the wave equation.
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Solution: We first compute both second derivatives that show up: ∂u = c sin(x − ct) ∂t ∂ 2u = −c2 cos(x − ct) ∂t2 ∂u = − sin(x − ct) ∂x ∂ 2u = − cos(x − ct) ∂x2 We can now verify that 2 ∂ 2u 2∂ u = c ∂t2 ∂x2
since this becomes −c2 cos(x − ct) = c2 (− cos(x − ct)) 7. (a) (5 points) Find the derivative of the function f : f (x, y) = (2xy, sin(xy)).
R2
−→
R2
given by
Solution: We can write out the matrix of the derivative: " Df =
∂f1 ∂x ∂f2 ∂x
∂f1 ∂y ∂f2 ∂y
#
"
2y 2x = y cos(xy) x cos(xy)
#
(b) (5 points) Let u = g(s, t) = s2 − t3 + 3. If s(x, y) = 2xy and t(x, y) = sin(xy), ∂u ∂u and . find ∂x ∂y Solution: From the chain rule, we know that D(g ◦ f ) = Dg · Df , so h
∂u ∂x
∂y ∂y
i
=
h
∂u ∂s
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∂u ∂t
i
" ·
∂f1 ∂x ∂f2 ∂x
∂f1 ∂y ∂f2 ∂y
#
so h
∂u ∂x
∂y ∂y
i
h
= 2s −3t2
i
"
# 2y 2x · y cos(xy) x cos(xy)
h i = 2s · 2y − 3t2 · y cos(xy) 2s · 2x − 3t2 · x cos(xy) So ∂u = 2s · 2y − 3t2 · y cos(xy) ∂x = 2(2xy) · 2y − 3 sin2 (xy) · y cos(xy) and ∂y = 2s · 2x − 3t2 · x cos(xy) ∂y = 2(2xy) · 2x − 3 sin2 (xy) · x cos(xy)
8. (12 points) Consider the function f (x, y) = (1 + xy)(x + y). Find the critical points, and determine whether each is a minimum, a maximum, or a saddle point. Solution: Firstly, we will find the critical points of f . For this we need: ∂f = y(x + y) + (1 + xy) = 1 + 2xy + y 2 ∂x ∂f = x(x + y) + (1 + xy) = 1 + 2xy + x2 ∂y Now these must both be zero, so we must solve 1 + 2xy + y 2 = 0 1 + 2xy + x2 = 0 We can subtract the equations to get the condition y 2 − x2 = 0, which means that y = ±x. Taking this and using it with the first equation, that becomes either 1 + 2x2 + x2 = 0 or 1 − 2x2 + x2 = 0, depending on the sign. But this first equation is 3x2 = −1, which is impossible, so we must have y = −x and the equation 1 − x2 = 0. The possible solutions for x are now ±1, and the critical points are (−1, 1) and (1, −1).
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Next we need to find the second order partial derivatives. ∂ 2f = 2y ∂x2 ∂f 2 = 2x + 2y ∂x∂y ∂f 2 = 2x ∂y 2 This means that the determinant of the Hessian matrix is 2y 2x + 2y = 4xy − (2x + 2y)2 = −4x2 − 4xy − 4y 2 2x + 2y 2x At our first critical point, (−1, 1), we get the value of −4, which means that we have a saddle point. At our second critical point, (1, −1), we also get the value of −4, so both critical points are saddle points, and there are neither relative minima nor maxima for this function. 9. (12 points) Find the minimum value of the function f (x, y, z) = x2 + y 2 + z 2 subject to the constraint that x + y + z = 12. Solution: The method of Lagrange Multipliers tells us that the minimum will occur when the derivatives of f and g(x, y, z) = x + y + z are scalar multiples. This means that we can solve the system of equations ∂f ∂x ∂f ∂y ∂f ∂z x+y+z
∂g ∂x ∂g =λ ∂y ∂g =λ ∂z = 12 =λ
We can compute the partial derivatives to get 2x = λ 2y = λ 2z = λ x + y + z = 12
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We see that λ 6= 0, or else we get x, y, and z are all 0, violating the final condition. Let’s first subtract the first two equations, giving 2x − 2y = 0 from which we see that y = x. Subtracting the first and third equations similarly gives z = x, so we know x = y = z. So the last equation becomes 3x = 12 so we get x = y = z = 4 is the unique minimum point. We can now find the minimum value that our function takes by evaluating: 42 + 42 + 42 = 48
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