Parametrics / Polars practice exam

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timer Asked: Dec 4th, 2017
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Question description

Do the problems in the attached study guide. Please show work. Can be handwritten or typed, I don't care.

Review for the Third Midterm of Math 150 B 12/05/2017 Problem 1 A curve has the parametric equations x = cos t, y = 1 + sin t, 0 ≤ t ≤ 2π dy a) Find dx when t = π4 . b) Find the equation of the line tangent to the curve at t = π/4. Write it in y = mx+b form. c) Eliminate the parameter t to find a cartesian (x, y) equation of the curve. d) Using (c), or otherwise identify the curve. Problem 2 Find the equation of the tangent line to the curve x = 3 + ln t, y = t2 + 2 at the point (3, 3). Problem 3 Consider the curve parametrically as x = t3 − t, y = t4 − 5t2 + 4. dy a) Find dx . b) Find all the points where tangent lines are horizontal. c) Find all the points where tangent line are vertical. d) Find the slope of the tangent line at the point on the graph of the curve when t = 1. Problem 4 a) Find the equations of the tangent lines to the cycloid x = rθ−r sin θ, y = r−r cos θ at θ = π2 . b) Find all the points on the graph of the cycloid at which the tangent line is either horizontal or vertical, and find the equations of the horizontal lines. Ans. a) y = x + r(2 − π2 ). b) equation of tangent line y = 2r. Problem 5 Consider the polar curve r = θ2 . a) Find parametric equations x = f (θ), y = g(θ) for this curve. b) Find the slope of the line tangent to this curve at the point (r, θ) = (π 2 , π). Problem 6 Find the slope of the tangent line to the curve that satisfies r = 1+edsin θ , at the point θ = π3 , where the equation is in polar coordinates with d > 0 and 0 < e < 1. Problem 7 √ a) Sketch the curve r = θ, in polar coordinates, for 0 ≤ θ ≤ 2π. Find the area enclosed by this curve. Problem 8 Find a polar equation for the circle x2 + (y − 3)2 = 9. Problem 9 Replace the following polar equations by equivalent Cartesian equations. a) r = 1 + 2r cos θ b) r = 1 − cosθ 4 c) r = 2 cos θ−sin . θ Problem 10 1 Find surface area of the solid generated by revolving the smooth curve C represented 3 2 by the parametric √ equations x(t) = 2t , y(t) = 3t , 0 ≤ t ≤ 1, about the x-axis. 24π Ans. S = 5 ( 2 + 1). Problem 11 I) Find the rectangular coordinates of each point whose polar coordinates are: a) (4, π3 ), b) (−2, 3π ). 4 II) Find Four polar coordinates (2 with r > 0 and 2 with r < 0) of each point whose rectangular coordinates √ are: a) (4, −4), b) (−1, − 3). Problem 12 Find the area of the region enclosed by the r = cos 2θ. Problem 13 Find the area of the region enclosed by the the limacon r = 2 + cos θ. Ans. 9π . 2 Problem 14 Find the area of the region that lies outside the cardioid r = 1 + cos θ and inside of the circle r = 3 cos θ. Ans. π. Problem 15 Find the tangent to the curve x = sec t, y = tan t − π/2 < t < π/2 at t = π/4. Problem 16 Find an equation of the sphere that passes through the point (4,3,-1) and has center (3,8,1). Problem 17 Find an equation of a sphere if one its diameters has endpoints (2,1,4) and (4,3,10). Problem 18 Show that the equation represents a sphere and find its center and radius. x2 + y 2 + z 2 − 2x − 4y + 8z = 15. Problem 19 Show that the equation represents a sphere and find its center and radius. 3x2 + 3y 2 + 3z 2 = 10 + 6y + 12z. Problem 20 Find a scalar projection and vector projection of ~b =< 1, 1, 2 > onto ~a =< −2, 3, 1 > . Problem 21 Find the work done by a force F~ = 8~i − 6~j + 9~k that moves an object from point (0,10,8) to the point (6,12,20) along a straight line. The distance is measured in meters and the force in newtons. Problem 22 A tow truck drags a stalled car along a road. The chain makes an angle of 30 degree with the road and the tension in the chain is 1500N . How much force is done by the truck in pulling the car 1 km? Problem 23 Find the angle between the vectors ~u =< 2, 2, −1 > and ~v =< 5, −3, 2 > . Problem 24 Find two unit vectors orthogonal to both vectors ~u =< 3, 2, 1 > and ~v =< −1, 1, 0 > . 2 Problem 25 Find a vector orthogonal to the plan through the points P (0, −2, 0), Q(4, 1, −2), R(5, 3, 1) and the area of triangle PQR. √ Ans. < 13, −14, 5 > and 12 390. Problem 26 Find the volume of the parallelepiped determined by the vectors ~a =< 1, 2, 3 >, ~b =< −1, 1, 2 >, ~c =< 2, 1, 4 > . Ans. 9 Problem 27 Graph and identify the curve defined by the parametric equations x = t2 , y = t + 1, −∞ < t < ∞. Problem 28 Find the tangent line to the curve x = et +t2 , y = e2t +3t at the point corresponding to t = 1. 2 +3 Ans. y − (e2 + 3) = 2ee+3 (x − (e + 1)). Problem 29 For the following parametric equations for the motion of a particle in the xy-plane x = 4 sin t, y = 5 cos t, 0 ≤ t ≤ 2π, Identify the particle’s path by finding a cartesian equation of it. Graph the Cartesian equation. Indicate the direction of the motion. Problem 30 Find parametric equations for the line through (-2,0,4) and parallel to ~v = 2~i+4~j −2~k. Problem 31 a) Find parametric equations for the line through P(-3,2,-3) and Q(1,-1,4). b) Find the distance between P and Q. Ans. a) x = −3 + 4t, y = 2 − 3t, z = −3 + 7t. Problem 32 Parametrize the line segment joining the points P(-3,2,-3) and Q(1,-1,4). Ans. x = −3 + 4t, y = 2 − 3t, z = −3 + 7t, 0 ≤ t ≤ 1. Problem 32 Consider the points (2,1,4) and (4,3,10). a) Express P~Q in terms of the unit vector ~i, ~j, ~k. b) Find the magntude of P~Q. c) Find the position vector of magnitude 6 in the direction of P~Q. Problem 33 A small rocket is launched from a table that is 3.36 ft above ground. Its initial velocity is 64f t/sec, and it is launched at an angle of 30 degree with respect to the ground. Air resistance is assumed to be negligeable, then its position after t seconds is x = 64 cos(30)t, y = 64 sin(30)t − 16t2 + 3.36. Find the rectangular equation that models its path. What type of path does the rocket follow? √ 1 Ans. y = − 192 x2 + 33 x + 3.36. 3

Tutor Answer

Barbartos
School: New York University

Complete.

150b Review
December 2017

Problem 1
a)
dy
=
dx

dy
dt
dx
dt

cos t
− sin t
= − cot t
=

At t =

π dy
4 , dx

= −cot

π
4



b)
From part a) we know the slope of the tangent line at t =
and y = 1 + √12 . Then solving for the y intercept,
1
1
1 + √ = (−1) √ + b
2
2

b=1+ 2
Then the tangent line is y = −x + (1 +



2)

c)
We proceed using the Pythagorean identity:
x2 = cos2 t
(y − 1)2 = sin2 t
x2 + (y − 1)2 = cos2 t + sin2 t = 1
Thus we have x2 + (y − 1)2 = 1
1

π
4 Att=π 4 ,

x=

√1
2

d)
The curve is a circle of radius 1 centered at (0, 1). We can recognize it from the
parametric equation since

Problem 2
The slope of the tangent line is given by the derivative.
dy
= 2t
dt
1 dy
dx
=
dt
t dx

=

dy
dt
dx
dt

= 2t2

dy
=2
At t = 1, (x, y) = (3, 3) so evaluating at t = 1, dx
Therefore, using the slope-intercept form, we have

3 = 2(3) − b =⇒ b = 3
So the tangent line is y = 2x − 3

Problem 3
a)
dy
= 4t3 − 10t
dt
dx
= 3t2 − 1
dt
dy
4t3 − 10t
=
dx
3t2 − 1

b)
The tangent lines are horizontal when the derivative is 0. This occurs when
4t3 − 10t = 0
2t(2t2 − 5) = 0
r
t = 0, ±

5
2

Substituting
q the equations for x and y, this yields (x, y) = (0, 4),
 q
  into
5
9
3
5
9
3
2
2, −4 , −2
2, −4 .

2

c)
The tangent lines are vertical when the denominator of the derivative is 0 since
it does not exist. Then 3t2 − 1 = 0 =⇒ t = ± √13 . Substituting into the


equations for x and y, we get (x, y) = ± √23 , 22
9

d)
Evaluating at t = 1, we have
dy
4(1)3 − 10(1)
=
= −3
dx
3(1)2 − 1

Problem 4
a)
dx
= r − r cos θ

dy
= r sin θ

dy
sin θ
=
dx
1 − cos θ
dy
At θ = π2 , dx
= 1 so the slope of the tangent line is 1.
At θ = π2 , (x, y) = πr
2 − r, r . Using the slope-intercept form, we have

 πr
−r +b
r=
2


Therefore, b = r 2 − π2 so the tangent line is y = x + r 2 − π2 .

b)
Tangent lines are vertical when the derivative does not exist and horizontal when
the derivative is 0. Thus, sin θ = 0 =⇒ θ = nπ for n ∈ Z. However, 1 − cos θ =
0 has root θ = 2nπ. Therefore, we must check the limit using l’Hopital’s rule.
dy
θ
= cos
Taking the derivative of the numerator and denomoinator, dx
sin θ . At θ =
2nπ, the denominator goes to 0 so the derivative is undefined. Therefore, there
is a vertical tangent at θ = 2nπ. Substituting in for the coordinates, we have
vertical ta...

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Anonymous
Wow this is really good.... didn't expect it. Sweet!!!!

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