Multivariable Calculus Practice Midterm 2

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Multivariable Calculus Practice Midterm 2. Prof. Fedorchuk. 1. (10 points) Compute the arc length of the curve r(t) = 〈 t,. 4. 3 t. 3. 2 ,t2. 〉 , on the interval 1 ≤ t ≤ 3.
Multivariable Calculus Practice Midterm 2 Prof. Fedorchuk

1.

(10 points) Compute the arc length of the curve   4 3 2 ~r(t) = t, t 2 , t , 3 on the interval 1 ≤ t ≤ 3.

D E √ 1 Solution: We have ~r 0 (t) = 1, 2t 2 , 2t and k~r 0 (t)k = 1 + 4t + 4t2 = 1 + 2t. Hence the arc length is Z 3 Z 3 3 0 k~r (t)kdt = (1 + 2t)dt = (t2 + t) = (12) − 2 = 10. 1

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2.

(10 points) Reparametrize the curve ~r(t) = het , et sin t, et cos ti with respect to arc length measured from the point (1, 0, 1) in the increasing direction of t.

Solution: We compute that ~r 0 (u) = heu , eu sin u + eu cos u, eu cos u − eu sin ui and that p k~r 0 (u)k = e2u + (eu sin u + eu cos u)2 + (eu cos u − eu sin u)2 q √ = e2u + 2e2u (sin2 (u) + cos2 (u)) = 3eu . The arc length function is √ 3eu du = 3(et − 1). 0 0 √ Solving for t in terms of s we obtain t = ln(1 + s/ 3). Hence the arc length parameterization of the curve is D   √ √ √ √ √  √ E p~(s) = ~r(ln(1+s/ 3)) = eln(1+s/ 3) , eln(1+s/ 3) sin ln(1 + s/ 3) , eln(1+s/ 3) cos ln(1 + s/ 3) D   √ √ √  √ √ E = 1 + s/ 3, (1 + s/ 3) sin ln(1 + s/ 3) , (1 + s/ 3) cos ln(1 + s/ 3) Z

s = s(t) =

t

0

k~r (u)kdu =

Z t√

(We plugged in the expression for t in terms of s into the original equation of the vector function ~r(t).

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(10 points) √ √ Find the tangent lines to the curves r1 (t) = ht, t2 , t3 i and r2 (u) = h1, 2 sin(u), 2 cos(u)i at the point (1, 1, 1).

Solution: Solving for t and u, we find that t = 1 and √ u = π/4 give√the point (1, 1, 1). We 0 2 0 also compute that r1 (t) = h1, 2t, 3t i and r2 (u) = h0, 2 cos(u), − 2 sin(u)i. The tangent line to r1 (t) at t = 1 has direction vector r01 (1) = h1, 2, 3i. Hence this tangent line has parametric equation   x = 1 + s y = 1 + 2s  z = 1 + 3s The tangent line to r2 (u) at u = π/4 has direction vector r02 (π/4) = h0, 1, −1i. Hence this tangent line has parametric equation   x = 1 y =1+s  z = 1 − s

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p x2 − y (10 points) Let f (x, y) = . 1+x a. (5 pts)

Find the partial derivatives ∂f /∂x and ∂f /∂y.

Solution: p 2x p (1 + x) − x2 − y 2 x2 − y ∂f x(1 + x) − (x2 − y) x−y p p = = = ∂x (1 + x)2 (1 + x)2 x2 − y (1 + x)2 x2 − y ∂f 1 =− p ∂y 2 x2 − y(1 + x)

b. (5 pts)

Describe (using inequalities) and sketch the domain of f (x, y).

Solution: The domain is defined by x 6= −1 and x2 − y ≥ 0, or ( x 6= −1 y ≤ x2 2

The domain is the region on or under the parabola y = x2 with the exception of the vertical line x = −1.

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(10 points) a. (5 pts) Find the limit x2 + y 2 (x,y)→(1,2) xy lim

Solution:

x2 + y 2 is defined at (1, 2), it is continuous at (1, 2). So the xy 12 + 22 5 limit is simply the value of the function at (1, 2), which is = . 2 2

Since the rational function

b. (5 pts)

Prove that the following limit does not exist: 3x2 − y 2 lim (x,y)→(0,0) xyex

Solution: Since lim(x,y)→(0,0) ex = e0 = 1, the limit in question exists only if the limit 3x2 − y 2 lim (x,y)→(0,0) xy exists. We proceed to show that the latter limit does not exist. Take a curve through (0, 0) given by x = y = t. Then along this curve, the function takes 3t2 − t2 values = 2. Take now a curve through (0, 0) given by x = t and y = −t. Along t2 3t2 − t2 = −2. Since these two values are distinct, this curve, the function takes values −t2 3x2 − y 2 we conclude that the limit of at (0, 0) does not exist. xy

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(10 points)

a. (5 pts) are zero.

Find all (x, y) such that the first partial derivatives of f (x, y) = xyey

Solution: We compute that fx = yey and fy = x(1 + y)ey . Setting these to zero we obtain a system ( yey = 0 x(1 + y)ey = 0 Since ey > 0, this gives ( y=0 x(1 + y) = 0 The only solution of this system is x = y = 0. Hence the answer is (0, 0). b. (5 pts)

Find all the second partial derivatives of f (x, y) = xyey .

Solution: We computed that fx = yey and fy = x(1 + y)ey . Therefore, differentiating once more gives: fxx = 0 fxy = (1 + y)ey fyy = x(2 + y)ey

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(5 points) a. (5 pts)

Let z = ln(u3 + v 4 ), where u = t4 and v = t2 . Compute dz/dt.

Solution: By the Chain Rule, dz ∂z du ∂z dv 3u2 4v 3 12t11 + 8t7 3 = + = 3 (4t ) + (2t) = . dt ∂u dt ∂v dt u + v4 u3 + v 4 t12 + t8

√ (15 points) A curve in space is given by the vector function ~r(t) = h2 sin(t), sin(t), 5 cos(t)i. Compute unit tangent vectors to the curve.

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√ 0 5 sin(t)i. It follows that k~r 0 (t)k = Solution: We compute that ~ r (t) = h2 cos(t), cos(t), − q √ √ (2 cos(t))2 + (cos(t))2 + ( 5 sin(t))2 = 5. Therefore, the unit tangent vector at time t is   ~r 0 (t) 2 1 T(t) = = √ cos(t), √ cos(t), − sin(t) k~r 0 (t)k 5 5

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(10 points) Answer the following questions. Give a brief explnation. Each question is worth 2 point. Question 1: Can two different level curves of a function intersect? Answer: No! Different level curves correspond to different values of the function. Since a function cannot take different values at a point, different level curves never intersect.

Question 2: What are the level surfaces of the function f (x, y, z) = 5x + 4y + 3z. Answer: These are planes 5x + 4y + 3z = c.

∂ 2f ∂ 2f and always equal? ∂x∂y ∂y∂x Answer: No! Mixed partials are equal for twice continuously differentiable functions (Clairaut’s theorem), but there are examples of functions whose mixed partials are not equal.

Question 3: Are the mixed partials

Question 4: Define what it means for a function f (x, y) to be continuous at a point (a, b). Answer: lim f (x, y) = f (a, b) (x,y)→(a,b)

Question 5: A space curve has a unique parameterization. Answer: No! Space curves have many different parameterizations. For example, a curve ~r(t), t ∈ [a, b], is also parameterized by ~r(2s), s ∈ [a/2, b/2].

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