Calculus Midterm Exam

Calculus Midterm Exam

1. (a) (6 points) Expand f (x) = ln(1 + x) in powers of x.

Solution: Since f (0) = 0, and f^(k)(x) = (−1)^k−1(k − 1)!

(x + 1)^k , f^(k)(0) = (−1)^k−1(k − 1)! for each k ≥ 1. f (x) =

∑

(−1)^k−1 k x^k.

(b) (6 points) Find a powers series representation for the improper integral Z _x

0

ln(1 + t)

t dt.

Solution: Using part (a), we haveln(1 + t) t =

∑

k=1

(−1)^k−1

k t^k−1. Therefore, Z _x

0

ln(1 + t) t dt = Z _x

0

∑

(−1)^k−1

k t^k−1dt =

∑

(−1)^k−1 k² x^k.

2. Let f (x) =e^x− 1 x .

(a) (5 points) Expand f (x) in powers of x.

Solution: Since d^ke^x

dx^k |_x=0 = e^x|_x=0 = 1 for each k ≥ 0, e^x =

∑

k=0

x^k

k!, and f (x) = e^x− 1

x =

∑^∞_k=1x^k−1 k! .

(b) (5 points) Show that

∑

n

(n + 1)! = 1.

Solution: Using part (a), we get f⁰(x) =

∑

k=1

(k − 1)x^k−2 k! =

∑

n=0

nxⁿ⁻¹

(n + 1)!=

∑

n=1

nxⁿ⁻¹

(n + 1)!, where we have set n = k − 1 in the second last equality. Now f⁰(x) = e^x

x −e^x− 1

x² , we get 1 = f⁰(1) =

∑

n (n + 1)!.

3. (10 points) Find an equation for the tangent plane and scalar parametric equations for the normal line to the surface z³+ xyz − 2 = 0 at (1, 1, 1).

Solution: Let f (x, y, z) = z³+ xyz − 2 = 0. Thus, ∇f = (yz, xz, xy + 3z²), and, at (1, 1, 1), ∇f = (1, 1, 4). Therefore, (x − 1) + (y − 1) + 4(z − 1) = 0 is an equation for the tangent plane, while (x, y, z) = (t + 1,t + 1, 4t + 1) are parametric equations for the normal line to the surface at (1, 1, 1).

Calculus Midterm Exam

4. (a) (6 points) Find the unit tangent vector and the principal unit normal vector for the curve r(t) = 2ti + lntj − t²k.

Solution: Since r⁰(t) =¡ 2,1

t, −2t¢

, kr⁰(t)k = 2t +1

t = 2t²+ 1

t , the unit tangent vector is T(t) = r⁰(t)

kr⁰(t)k=

³ 2t

2t²+ 1, 1

2t²+ 1, − 2t² 2t²+ 1

´

=

³ 2t

2t²+ 1, 1

2t²+ 1, 1 2t²+ 1−1

´

. Furthermore,

T⁰(t) =

³ 2

2t²+ 1 − 8t²

(2t²+ 1)², − 4t

(2t²+ 1)², − 4t (2t²+ 1)²

´

= 2

(2t²+ 1)²

³

1 − 2t², −2t, −2t

´ ,

and kT⁰(t)k = 2

(2t²+ 1), the principal unit normal vector is N(t) = T⁰(t) kT⁰(t)k

= 1

(2t²+ 1)

³

1 − 2t², −2t, −2t

´ .

(b) (6 points) Find the length of the plane curve y = x^3/2for 0 ≤ x ≤ 4.

Solution: The length is Z ₄

0

q

1 + (dy/dx)²dx = Z ₄

0

r 1 +9x

4 dx = 8

27(1 +9x 4 )^3/2|⁴₀

= 8 27

h

10^3/2− 1 i

.

5. (a) (6 points) Find the directional derivative of f (x, y) = e^2x(cos y − sin y) at (¹₂, −^π₂) in the direction of the greatest increase of f .

Solution: Since the directional derivative D_uf (p) of f in the direction u at a point p satisfies that Duf (p) = k∇f (p)k cosθ, whereθ ∈ [0,π] is the angle between u and∇f (p). Therefore, Duf (p) reaches maximum value k∇f (p)k whenθ = 0, or u =_k^∇_∇^{f (p)}_{f (p)k}.

Since∇f =¡

2e^2x(cos y − sin y), −e^2x(sin y − cos y)¢

,∇f (¹₂, −^π₂) =¡ 2e, e¢

, and the directional derivative of f at (¹₂, −^π₂) in the direction ¹₅(2, e) increases fastest with the rate equals to k∇f (¹₂, −^π₂)k =√

5e.

(b) (6 points) Find the directional derivative of g(x, y, z) = sin(xyz) at (¹₂,¹₃,π) in the direction of the greatest decrease of g.

Solution: From part (a), we know that the directional derivative D_ug(p) = k∇g(p)k cosθ^, reaches minimum value −k∇g(p)k whenθ ⁼π^{, or u =} _k⁻_∇^∇g(p)_g(p)k^.

Since ∇g =¡

yz cos(xyz), xz cos(xyz), xy cos(xyz)¢

, ∇g(¹₂,¹₃,π) =

√3 2

¡_π

3,^π₂,¹₆¢

, and the direc- tional derivative of g at (¹₂,¹₃,π) decrease fastest in the direction ^{√ −1}

13π²+1

¡2π^{, 3}π^{, 1¢with the}

rate −k∇g(¹₂,¹₃,π^{)k = −}

√39π^{2+ 3}

12 .

6. (10 points) Find the curvature of the plane curve x(t) = 2e^−t, y(t) = e^−2t.

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Calculus Midterm Exam

Solution: Let r(t) =¡

x(t), y(t)¢

=¡

2e^−t, e^−2t¢

. Then r⁰(t) = −2e^−2t¡ e^t, 1¢

, and ds

dt = kr⁰(t)k = 2e^−2t

p

e^2t+ 1. Thus,

the unit tangent vector T (t) = r⁰(t)

kr⁰(t)k= − 1

√e^2t+ 1

¡e^t, 1¢

=¡

− e^t

√e^2t+ 1, − 1

√e^2t+ 1

¢.

T⁰(t) =¡

− e^t

√

e^2t+ 1+ e^3t

(e^2t+ 1)^3/2, e^2t (e^2t+ 1)^3/2

¢= e^t (e^2t+ 1)^3/2

¡− 1, e^t¢ ,

the principal unit normal vector N(t) = T⁰(t) kT⁰(t)k =¡

− 1

√e^2t+ 1, e^t

√e^2t+ 1

¢,

and dN

dt = e^t (e^2t+ 1)^3/2

¡e^t, 1¢

. The curvatureκ= kdN

dsk = kdN dt k

.ds dt

= e^t

(e^2t+ 1)^3/2 p

e^2t+ 1 1 2e^−2t√

e^2t+ 1 = e^3t 2(e^2t+ 1)^3/2.

7. (a) (5 points) A set of points is said to be convex provided that every pair of points in the set can be joined by a line segment that lies entirely within the set. Show that, if k∇f (x)k ≤ M for all x in some convex setΩ, then | f (x1) − f (x₂)| ≤ Mkx₁− x₂k for all x₁and x₂inΩ.

Solution: For any x₁, x₂∈Ω, the line segment r(t) = (1−t)x1+tx₂lies entirely inΩand joins x₁= r(0) to x₂= r(1). Thus, the function g(t) = f (r(t)) is differentiable for t ∈ [0, 1]. Mean Value Theorem implies that there exists a t0∈ (0, 1) such that | f (x₂)− f (x₁)| = |g(1)−g(0)| =

|g⁰(t₀)(1 − 0)| = |∇f (r(t₀)) · r⁰(t₀)| = |∇f (c) · (x₂− x₁)| ≤ k∇f (c)k · kx₂− x₁k ≤ Mkx₂− x₁k, where c = r(t₀).

(b) (5 points) Show that if f is differentiable at each point of the line segment ab and f (a) = f (b), then there exists a point c between a and b for which∇f (c) ⊥ (b − a).

Solution: Let r(t) = (1 − t)a + tb for t ∈ [0, 1], and set g(t) = f (r(t)). Mean Value Theorem implies that there exists t₀∈ (0, 1) such that 0 = f (b) − f (a) = g(1) − g(0) = g⁰(t₀)(1 − 0) =

∇f (r(t0)) · r⁰(t0) =∇f (c) · (b − a), where c = r(t0) ∈ ab. Thus, we have∇f (c) ⊥ (b − a).

8. Let t, n, b = t × n denote the unit tangent, principal normal, and binormal at each point of a regular curve.

(a) (4 points) Show that db/ds ⊥ b.

Solution: Since hb, bi(s) = 1 for each s, we have 0 = d

ds1 = d

dshb, bi = 2hdb

ds, bi. Thus, db

ds ⊥ b.

(b) (4 points) Use the fact thatdb ds = d

ds(t × n) to show that db/ds ⊥ t.

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Calculus Midterm Exam

Solution: Since db ds = d

ds(t × n) = dt

ds× n + t ×dn

ds =κn × n + t ×dn

ds = t ×dn

ds, we have hdb

ds, ti = ht ×dn

ds, ti = 0 which implies that db ds ⊥ t.

(c) (4 points) Show that db

ds =τn, for some scalarτ^.

Solution: Since t, n, b form an orthonormal basis of R³ at each r(s) in the curve, and db ds is perpendicular to both b and t from parts (a) and (b), there exists a scalarτ^{such that db}

ds =τ^n.

9. Let f (x, y) = x²+ y²− 2x + 2y + 2 and D = {(x, y) | x²+ y²≤ 8}.

(a) (6 points) Find local extreme values of f in the interior of D, i.e. inside {(x, y) | x²+ y²< 8};

determine the local maxima and minima and the saddle points.

Solution: ∇f = (2x − 2, 2y + 2) = (0, 0) implies that (x, y) = (1, −1). At (1, −1), the Hes- sian matrix Hess( f ) of f ,

µf_xx f_xy f_yx f_yy

¶

=

µ2 0 0 2

¶

is positive definite, f (1, −1) = 0 is a local minimum.

(b) (6 points) Use the method of Lagrange to determine the extreme values of f on the boundary of D, i.e. on the circle {(x, y) | x²+ y²= 8}.

Solution: Let g(x, y) = x²+ y²− 8 = 0 be the constraint condition. Solve λ, x, y such that

∇f (x, y) =λ∇g(x, y), which is equivalent to the system (λ − 1)x = −1 and (λ^{− 1)y = 1.}

Sinceλ−1 6= 0, we get (x, y) =¡

− 1

λ^{− 1,} 1 λ^{− 1}

¢. Substituting these into the constraint g = 0, we get (x, y) = (−2, 2), or (x, y) = (2, −2), and f (−2, 2) = 18 is the maximum value, while

f (2, −2) = 2 is the minimum value of f on the circle {(x, y) | x²+ y²= 8}.

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