Download MATH 2210-1 Exam 1 Solutions: Parametric Curves, Velocity, Acceleration, and Planes and more Exams Advanced Calculus in PDF only on Docsity!
MATH 2210-
EXAM 1 SOLUTIONS
June 7 2006
- Given is a parametric representation of a curve
x = cos t + 1, y = sin 2 t, 0 ≤ t ≤ π/2.
(a) Eliminate the parameter.
(x − 1)^2 = cos^2 t
(x − 1)^2 + y = 1
or y = 1 − (x − 1)^2
(b) What curve is the given curve part of?
Parabola.
- Given is a parametric representation of the curve x = 1 + sin t, y = t,
z = 1 − cos t, 0 ≤ t ≤ 10.
(a) Find the length of the curve.
x′(t) = cos t
y′(t) = 1
z′(t) = sin t
L =
∫ (^10) 0
cos^2 t + 1 + sin 2 tdt =
∫ (^10) 0
2 dt = 10
(b) Find the symmetric equations of the tangent line to the curve at t = π/4.
x ′ (π/4) = 1/
y′(π/4) = 1
z′(π/4) = 1/
The direction vector is < 1 /
2 > (and also < 1 ,
x(π/4) = 1 + 1/
y(π/4) = π/ 4
z(π/4) = 1 − 1 /
The symmetric equation is
x − 1 − 1 /
y − π/ 4 √ 2
z − 1 + 1/
- Let x = cos t, y = t 2 + 1, t 6 = nπ. Find dy/dx without eliminating the parameter.
dy/dt = 2t
dx/dt = − sin t
Since dx/dt 6 = 0, dy/dx =
dy/dt
dx/dt
2 t
sin t
- Nina wants to swim from one bank of a 1 mile wide river to the point
directly across. She can swim at a speed of 2 miles per hour. The speed of the current is 0.2 miles per hour. In what direction should she swim, and how long will it take her to swim across? (Draw the picture).
Nina should swim in the direction of a vector < x, 0. 2 >.
Her speed with respect to the water is the length of the vector, hence x^2 = 2^2 − 0. 22
Her speed with respect to the ground is x. Therefore, it will take her 1 /
3 .96 hours to swim across.
- Find the center and the radius of the sphere whose equation is
x 2 − 6 x + y 2 − 4 y + z 2
Complete the square:
x 2 − 6 x + 9 + y 2 − 4 y + 4 + z 2
(x − 3)^2 + (y − 2)^2 + (z + 4)^2 = 25
The center is (3, 2 , −4) and the radius is 5.
- Let ~r(t) = cos t~i + sin t~j + t^2 ~k. At t = 0, find
(a) velocity vector
~v(t) = ~r ′ (t) = − sin t~i + cos t~j + 2t~k
~v(0) = ~j
Let Q = (1, 2 , −1).
Pick a point P in the plane, for instance, P = (0, 0 , 3).
Then
P Q =< 1 , 2 , − 4 >
d = |−→pr< 2 ,− 1 , 1 > < 1 , 2 , − 4 > | =
∣ ∣ ∣ ∣ ∣
< 2 , − 1 , 1 >^2
- Find the angle between the planes 2x − y + 5z = 0 and x + 3y + z = 12.
cos θ =
θ = cos − (^1) √ 4 30
√ 11
- Let A = (0, 1 , −1), B = (2, 0 , 3) and C = (2, 1 , 0).
(a) Find the area of the triangle ABC −→ AC =< 2 , 0 , 1 >,
BC =< 0 , 1 , − 3 >
area=1/ 2 |
AC ×
BC| = 1/ 2 | < − 1 , 6 , 2 > | =
(b) Find an equation of the plane through A, B and C.
The normal is
AC ×
BC =< − 1 , 6 , 2 >
An equation is −x + 6(y − 1) + 2(z + 1) = 0.
