An object is propelled upward at an angle θ, 45^∘<θ<90^∘, to the horizontal with an initial velo

An object is propelled upward at an angle θ, 45^∘<θ<90^∘, to...

An object is propelled upward at an angle $\theta, 45^{\circ}<\theta<90^{\circ},$ to the horizontal with an initial velocity of $v_{0}$ feet per second from the base of a plane that makes an angle of $45^{\circ}$ with the horizontal. See the illustration. If air resistance is ignored, the distance $R$ that it travels up the inclined plane is given by the function $R(\theta)=\frac{v_{0}^{2} \sqrt{2}}{16} \cos \theta(\sin \theta-\cos \theta)$ (a) Show that $$ R(\theta)=\frac{v_{0}^{2} \sqrt{2}}{32}[\sin (2 \theta)-\cos (2 \theta)-1] $$ (b) In calculus, you will be asked to find the angle $\theta$ that maximizes $R$ by solving the equation $$ \sin (2 \theta)+\cos (2 \theta)=0 $$ Solve this equation for $\theta$ (c) What is the maximum distance $R$ if $v_{0}=32$ feet per second? (d) Graph $R=R(\theta), 45^{\circ} \leq \theta \leq 90^{\circ}$, and find the angle $\theta$ that maximizes the distance $R$. Also find the maximum distance. Use $v_{0}=32$ feet per second. Compare the results with the answers found in parts (b) and (c). (IMAGE CAN'T COPY)

An object is propelled upward at an angle $\theta, 45^{\circ}<\theta<90^{\circ},$ to the horizontal with an initial velocity of $v_{0}$ feet per second from the base of a plane that makes an angle of $45^{\circ}$ with the horizontal. See the illustration. If air resistance is ignored, the distance $R$ that it travels up the inclined plane is given by the function
$R(\theta)=\frac{v_{0}^{2} \sqrt{2}}{16} \cos \theta(\sin \theta-\cos \theta)$
(a) Show that
$$
R(\theta)=\frac{v_{0}^{2} \sqrt{2}}{32}[\sin (2 \theta)-\cos (2 \theta)-1]
$$
(b) In calculus, you will be asked to find the angle $\theta$ that
maximizes $R$ by solving the equation
$$
\sin (2 \theta)+\cos (2 \theta)=0
$$
Solve this equation for $\theta$
(c) What is the maximum distance $R$ if $v_{0}=32$ feet per
second?
(d) Graph $R=R(\theta), 45^{\circ} \leq \theta \leq 90^{\circ}$, and find the angle $\theta$
that maximizes the distance $R$. Also find the maximum
distance. Use $v_{0}=32$ feet per second. Compare the
results with the answers found in parts (b) and (c).
(IMAGE CAN'T COPY)

Key Concepts

Graphical Analysis

Graphical analysis involves plotting a function over a specified domain to visually inspect its behavior, including identifying local and global maxima or minima. This method supplements analytical approaches, offering a clear picture of the function's structure and helping to verify the results obtained by solving equations or using calculus. Graphing can also aid in understanding how changes in parameters, like the launch angle, influence the outcome.

Calculus Optimization

Calculus optimization involves finding maximum or minimum values of a function by differentiating the function, setting the derivative equal to zero, and solving for the variable. This method is widely used in physics and engineering to determine optimal parameters, such as the angle of launch that maximizes the distance traveled by a projectile. It includes verifying that the obtained critical points indeed correspond to a maximum (or minimum), often using second derivative tests or analyzing the behavior of the function.

Solving Trigonometric Equations

Solving trigonometric equations involves using trigonometric identities, algebraic manipulation, and the periodic properties of trigonometric functions to isolate the variable, often representing an angle. This process is commonly required in optimization problems where an equation derived from setting a derivative equal to zero must be solved to find critical points that maximize or minimize the function, such as determining the optimal launch angle for maximum distance.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the variable where both sides of the equality are defined. They are used to simplify expressions and rewrite functions in alternative forms, such as converting a combination of sine and cosine functions into a single trigonometric expression or into a sum or difference of angles. These identities are essential in tasks that involve the simplification and manipulation of expressions before further analysis or differentiation.

Projectile Motion

Projectile motion is a type of motion experienced by an object that is launched into the air and is subject only to gravity (with negligible air resistance). In problems involving inclined planes, the motion is decomposed into components parallel and perpendicular to the plane, which affects the trajectory and distance traveled. Understanding the kinematics of projectile motion is crucial for analyzing how initial velocity and launch angle determine the path of the projectile.

Transcript

00:01 Okay, in this problem we have an object, which is this little bomb being thrown or propelled somehow up this hill.

00:12 The hill is 45 degrees here, and it's making an angle theta with the horizontal, and its velocity is v -0, okay, which just means initial velocity, feet per second.

00:29 And the distance it travels up the hill is given by this formula right here r equals okay so the first thing they want us to do is show that that r is equal to this formula right here so the point here is we're practicing using double angled formulas so i'm going to start with this second one and turn it into the first one so i'm going to start here and i'm going to get to here kind of did that backwards but that's okay.

01:02 I'll start with the red.

01:06 So v0 squared, square root of 2 over 32 times the sine of 2 theta minus the cosine of 2 theta minus 1 is equal to v0 squared, square to 2 over 32.

01:21 Can place the sine 2 theta.

01:23 I'm going to put its identity, which is 2, sine theta, cosine theta, minus.

01:30 Okay, now i want this cosine 2 theta to get out of here, but remember there's three identities for it so i look to see what it is i'm trying to get i'm trying to get a cosine times sign which is going to be this i'm also trying to get a cosine times cosine okay so that tells me to use the identity that has only the cosine in it and it goes to cosine squared theta minus one and then we have this minus one right there all right so then the first thing you should notice is that this minus one and that minus one are going to cancel so i'm just going to do that because this one's going to be negative minus one minus one.

02:13 Okay, so those are gone.

02:16 Okay, now both of these have a two in it, so let's factor that out.

02:22 So this equals v0 squared, square to 2, 32 times 2.

02:28 Oh, they both have a cosine.

02:29 Let's take two cosine out.

02:32 And what's left is sine theta minus cosine theta.

02:37 And then we can cancel this two with this 32.

02:39 So that leaves us v0 squared or times the square to 2 over 16 times the cosine of theta times the sine of theta minus the cosine of theta, which is what we're trying to get the green one.

02:57 Next, we want to solve this equation, sine 2 theta plus cosine 2 theta equals 0.

03:05 Okay, the reason we're trying to solve it is because we're trying to find the maximum r value here in this part of the process of solving that.

03:15 Okay, so we've got sine 2 theta plus cosine 2 theta equals 0.

03:23 Okay, i can put identities in, but instead what i'm going to do is i'm going to move one on one side and one on the other, and then i'm going to square both sides, because that's how i know the relationship between sign and cosine.

03:41 Okay, so i'm going to square both sides.

03:43 The problem with squaring both sides is when i get answers, i'm going to have to check them, because squaring sometimes brings extraneous solutions.

03:54 Okay, so i get cosine squared 2 theta.

03:57 All right, i'm going to take the cosine squared out, and i'm going to put 1 minus sign squared in its place.

04:03 So i have sine squared 2 theta equals to 1 minus sine squared 2 theta.

04:11 Add that to both sides, so 2, sine square, 2 theta equals 1.

04:18 Sign squared 2 theta equals 1 half take the square root of both sides and don't forget plus and minus so i get the sign of 2 theta equals plus and minus 1 over the square root of 2 that's awesome because i know the answer there the angle whose sign is one over the square root of 2 is pi over 4 okay but i got plus and minus so it's pi over 4 in all quadrants pi over 4 3 pi over 4 5 pi over 4 7 5 pi over 4 7 pi over 4 7 pi over 4 7 pi over 4 plus because i've got a two theta here i have to go around again 9 pi over 4 11 pi over 4 13 pi over 4 and 15 pi over 4 and then to find theta divide everything by 2 so pi over 8 3 pi over 8 5 pi over 8 7 pi over 8 9 pi over 8 9 pi over 8 11 pi over 8 13 pi over 8 15 pi over 8.

05:31 Now remember i said because i squared both sides, i got to check my answer.

05:35 So let's start with pi over 8...

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