Block #3: Space Curves, Inverse Trig Functions Goals: …apsc171/Notes/notes03.pdf · 2019-11-14 · Cosine and arccosine - 1 Cosine and arccosine The inverse of cosine is obtained

Block #3: Space Curves, Inverse Trig Functions

Goals:

•Applications to Motion Problems• Inverse Trigonometric Functions

1

Motion Over Time - Taxi Problem - 1

Motion Over Time - Taxi Problem

Problem. Suppose you are a passenger in the back seat of a taxithat is speeding along so that its location at time t (seconds) is r(t) =〈100t2, 10t〉, measured in meters.What is the shape of the path of the taxi?


r(t) = 〈100t2, 10t〉

Problem. What is your acceleration at time t = 0?


r(t) = 〈100t2, 10t〉

Problem.Given that your mass is 70 kilograms, and assuming thatyou forgot to put on your seat belt, with what force is the car doorpressing against you at the instant t = 0?


Problem. In the same taxi scenario, as t increases from 0 to 10,which of the following alternatives best describes what happens?

A. The force of the door against your body stays the same.

B. The force of the door against your body decreases, but the forceof the back of your seat against your body increases.

C. The force of the door against your side remains the same, but youfind yourself sliding forward towards the back of the seat in frontof you.

Acceleration in Three Dimensions - 1

Acceleration in Three Dimensions

The calculation of acceleration in three-dimensional space is exactlythe same as in two dimensions, for exactly the same reasons. To findthe acceleration all you have to do is differentiate each of the (three)components twice:

If the position of a moving particle is given by

r(t) = 〈x(t), y (t), z (t)〉then its acceleration is given by

r′′(t) = 〈x′′(t), y′′(t), z′′(t)〉.


The acceleration due to gravity is 9.8 meters per second per second.Suppose that between times t = −10 and t = 10, measured inseconds, a stunt plane follows the path

r(t) = 〈200t, 5t3, 800− 5t2〉without going upside down or banking significantly.

Problem. Will the pilot ever be lifted off his seat?

Notice that in this problem, and in all cases unless indi-cated otherwise, the third component (the z-component)is assumed to measure vertical distance.


(Example Continued) r(t) = 〈200t, 5t3, 800− 5t2〉

Visualizing Plane and Space Curves - 1

Visualizing Plane and Space Curves

Problem. Describe the path traced out by the parametric planecurve

r(t) = 〈2 cos(t), sin(t)〉 .

Visualizing Plane and Space Curves - 2

Problem. Given the answer to the preceding question, describe thefollowing parametric space curve:

r(t) = 〈2 cos(t), sin(t), t〉 .A. It is an ellipse tilted in the direction of the x-axis.

B. It is an ellipse tilted in the direction of the y-axis.

C. It is a helix.

D. It is a curve in the plane that spirals outward.

Spaceship Problem - 1

Spaceship Problem

We now complete this first unit by revisiting the problem posed atits beginning.Suppose a spaceship moves in three-dimensional space so that at timet ≥ 0 its coordinates are given as

r(t) = 〈cos(t), sin(t), e−t〉.


Problem. What does that tell us about the way this object moves?Try focusing on the xy coordinates and the z coordinates separatelyto simplify the problem.

r(t) = 〈cos(t), sin(t), e−t〉.


.r(t) = 〈cos(t), sin(t), e−t〉, t ≥ 0

Problem. What is the spaceship’s speed at time t = 0?


r(t) = 〈cos(t), sin(t), e−t〉, t ≥ 0

Problem. When will the spaceship experience the greatest acceler-ation?


r(t) = 〈cos(t), sin(t), e−t〉, t ≥ 0

Problem. If you were to fall off at time t = 1, what trajectory wouldyou follow? (Assume we are in space, with effectively zero gravity.)


Problem. Something to challenge you: How far from the origin willyou “land” on the (x, y)-plane if you fall off the spacecraft at time t?

Projectile Motion Using Vectors - 1

Projectile Motion Using Vectors

To finish our discussion of parametric curves, we turn to a problemthat examines the path of an object moving under the influence ofthe force of gravity. It may be a problem that you have seen in someversion already. Here we will emphasize vector notation in thesolution.


A projectile is fired from the origin, with angle of elevation α (radians)and speed 60 meters per second.

α

Assume that air resistance is negligible and that the only force on theprojectile is gravity (producing a downward acceleration of g = 9.8meters per second per second).

Problem. Draw a free-body diagram for the projectile.


α

Problem. Find a (vector-valued) formula for the position of theprojectile at time t.


α


r(t) =

⟨60 cos(α)t,

−9.8

2t2 + 60 sin(α)t

⟩Problem. If launched at angle α, how far down range will the pro-jectile land?


(Range continued) r(t) =

⟨60 cos(α)t,

−9.8

2t2 + 60 sin(α)t

⟩


Problem.At what angle α should the projectile be launched so thatit will land 300 meters down range?

Useful trig identity: sin(2α) = 2 cos(α) sin(α)

Inverse Trigonometric Functions - Introduction - 1

Inverse Trigonometric Functions

At the end of our earlier vector-based trajectory problem, we neededa way to find the angle θ when sin(θ) was given. That is, we foundwe needed an inverse for the sine function. Most of you will haveseen something about inverse trigonometric functions in high school,and you will know that when you want to know the angle whose sinehas a particular given value, you use the “SHIFT + sin” or “sin−1”button combination on a calculator to find that angle.

Inverse Trigonometric Functions - Introduction - 2

Problem. Why should you (as mathematicians) be suspicious ofsuch an easy implementation of the inverse of the sine function?

How can we remove the obstacle to an inverse of sine? (Clearly, theremust be a way since the calculator is doing something!)

Sine and arcsine - 1

Sine and arcsine

For convenience we call this newfunction Sin (x), where

Sin (x) = sin(x)

provided −π2 ≤ x ≤ π2 .

1

−1

π

2−

π

2

f(x) = Sin(x)

b

b


Problem. Sin(x) has an inverse:what are two notations for this in-verse function?

1−1

π

2

−

π

2

f(x) = arcsin(x)

b

b

The domain of arcsin is: The range of arcsin is:


Note that, as always, the graph can be produced using MATLAB:

x = linspace(-1, 1);

y = asin(x);

plot(x, y);

axis equal

Sine and Arcsine as Inverses - 1

Sine and Arcsine as Inverses

Since arcsin undoes what sin does, and vice-versa, the following equa-tions are true, but only for the specified values of x:

arcsin(sinx) = x, for − π

2≤ x ≤ π

2sin(arcsinx) = x, for − 1 ≤ x ≤ 1.

Problem. What is the value of arcsin(0.5)?

Sine and Arcsine as Inverses - 2

Problem. sin(−7π/5) = 0.951, so what is the value of arcsin(0.951)?

Cosine and arccosine - 1

Cosine and arccosine

The inverse of cosine is obtained by a calculation similar to the waythe inverse of sine was determined. We analyze cosine from 0 to π;this is shown in the graph on the right.For convenience, we could call this new function Cos(x) where

Cos(x) = cos(x)

provided 0 ≤ x ≤ π.Cos(x) satisfies the horizontal line test and therefore has an inversefunction which we call the inverse cosine function and denote itas

cos−1(x) or arccos(x)

noting that

cos−1 x 6= 1

cosx.


0

1

−1

π

2

π

f(x) = Cos(x)

b

b

0 1−1

π

2

π

f(x) = arccos(x)

b

b

The domain of arccos is: The range of arccos is:


Problem. When you enter arccos(2) (via the “cos−1” button) onyour calculator, it objects. Why is that?

A. The numbers involved are too large for the calculator to handle.

B. The calculator does not understand this business of taking the in-verse using only part of the cosine function.

C. The cosine function does not really have an inverse.

D. The number 2 is outside the domain of the function arccos.

Tan and arctan - 1

Tan and arctan

The inverse of tan is determined in the same way, only analyzing itfrom −π2 to π

2 . This is shown in the graph on the next page:As done before, we name this portion of the tan function Tan(x),where

Tan(x) = tanx

provided −π2 < x < π2 .

Tan(x) satisfies the horizontal line test and therefore has an inverse,which we call the inverse tangent function and denote it as

tan−1 x or arctan(x)

once again noting that

tan−1 x 6= 1

tanx.

Tan and arctan - 2

π

2−

π

2

f(x) = Tan(x)

π

2

−

π

2

f(x) = arctan(x)

The domain of arctan is: The range of arctan is:

What is the value of arctan(1)?

Missile Launch Problem - 1

Missile Launch Problem

For an example of where getting the domain of an inverse trig functionis relevant, consider the earlier example of an object moving in theelliptical orbit defined by

r(t) = 〈2 cos(t), sin(t)〉If we imagine a missile being carried around in this orbit, we canrelease the missile at any time and it would then proceed in a straightline (Newton’s first law).

Problem. At what time(s) could the missile be released so that itwill hit a target at (x, y) = (10, 0)?


(Example continued) r(t) = 〈2 cos(t), sin(t)〉, target at (10, 0).


(Example continued) r(t) = 〈2 cos(t), sin(t)〉, target at (10, 0).


Next week, we will look at how we can differentiate these inverse trigfunctions, and explore further applications of derivatives.

Documents

Block #3: Space Curves, Inverse Trig Functions Goals: …apsc171/Notes/notes03.pdf · 2019-11-14 · Cosine and arccosine - 1 Cosine and arccosine The inverse of cosine is obtained