Question
Which is the polar form of the parametric equations x = 4t and y equals t squared?
(1 point)
Responses
r = 16 tan θ sec θ
r = 16 tan θ sec θ
r equals 16 tangent squared theta
Image with alt text: r equals 16 tangent squared theta
r equals 16 secant squared theta
Image with alt text: r equals 16 secant squared theta
r = 16 sec θ
A bowling ball is rolled down the alley with a constant velocity of 1.5 m/sec. at an angle of 87° to the starting line. The position of the person throwing the bowling ball can be represented by the point left parenthesis 0 comma 0 right parenthesis. Where is the ball after 7 seconds?
(1 point)
Responses
left parenthesis 0.5 comma 10.5 right parenthesis
Image with alt text: left parenthesis 0.5 comma 10.5 right parenthesis
left parenthesis 0.1 comma 1.5 right parenthesis
Image with alt text: left parenthesis 0.1 comma 1.5 right parenthesis
left parenthesis 10.5 comma 10.5 right parenthesis
Image with alt text: left parenthesis 10.5 comma 10.5 right parenthesis
left parenthesis 29 comma 200 right parenthesis
A ball is thrown with a slingshot at a velocity of 110 ft./sec. at an angle 20° above the ground from a height of 4.5 ft. Approximately how long does it take for the ball to hit the ground? Acceleration due to gravity is 32 feet per second squared.
(1 point)
Responses
2.35 seconds
2.35 seconds
2.47 seconds
2.47 seconds
6.46 seconds
6.46 seconds
6.50 seconds
(1 point)
Responses
r = 16 tan θ sec θ
r = 16 tan θ sec θ
r equals 16 tangent squared theta
Image with alt text: r equals 16 tangent squared theta
r equals 16 secant squared theta
Image with alt text: r equals 16 secant squared theta
r = 16 sec θ
A bowling ball is rolled down the alley with a constant velocity of 1.5 m/sec. at an angle of 87° to the starting line. The position of the person throwing the bowling ball can be represented by the point left parenthesis 0 comma 0 right parenthesis. Where is the ball after 7 seconds?
(1 point)
Responses
left parenthesis 0.5 comma 10.5 right parenthesis
Image with alt text: left parenthesis 0.5 comma 10.5 right parenthesis
left parenthesis 0.1 comma 1.5 right parenthesis
Image with alt text: left parenthesis 0.1 comma 1.5 right parenthesis
left parenthesis 10.5 comma 10.5 right parenthesis
Image with alt text: left parenthesis 10.5 comma 10.5 right parenthesis
left parenthesis 29 comma 200 right parenthesis
A ball is thrown with a slingshot at a velocity of 110 ft./sec. at an angle 20° above the ground from a height of 4.5 ft. Approximately how long does it take for the ball to hit the ground? Acceleration due to gravity is 32 feet per second squared.
(1 point)
Responses
2.35 seconds
2.35 seconds
2.47 seconds
2.47 seconds
6.46 seconds
6.46 seconds
6.50 seconds
Answers
GPT-4o mini
Let's handle each part of your question step-by-step.
### 1. Polar Form of Parametric Equations
The parametric equations given are:
- \( x = 4t \)
- \( y = t^2 \)
First we need to express \( y \) in terms of \( x \):
- From \( x = 4t \), we can solve for \( t \):
\[
t = \frac{x}{4}
\]
Now substitute this expression for \( t \) into the equation for \( y \):
- \( y = t^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \)
We also have the relation between Cartesian coordinates and polar coordinates:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
Since \( y = \frac{x^2}{16} \), we can express it in terms of \( r \) and \( \theta \):
- \( r \sin \theta = \frac{(r \cos \theta)^2}{16} \)
- \( r \sin \theta = \frac{r^2 \cos^2 \theta}{16} \)
Assuming \( r \neq 0 \), we can divide both sides by \( r \):
- \( \sin \theta = \frac{r \cos^2 \theta}{16} \)
- Rearranging gives us:
\[
r = 16 \frac{\sin \theta}{\cos^2 \theta}
= 16 \tan \theta \sec \theta
\]
So the answer to the first part is:
**\( r = 16 \tan \theta \sec \theta \)**
### 2. Position of the Bowling Ball
The bowling ball is thrown with a velocity of 1.5 m/sec at an angle of 87°:
- The horizontal and vertical components of the velocity can be calculated as follows:
- \( v_x = v \cos(\theta) = 1.5 \cos(87°) \)
- \( v_y = v \sin(\theta) = 1.5 \sin(87°) \)
Using \( \cos(87°) \approx 0.0393 \) and \( \sin(87°) \approx 0.998 \):
- \( v_x \approx 1.5 \times 0.0393 \approx 0.059 \, \text{m/sec} \)
- \( v_y \approx 1.5 \times 0.998 \approx 1.497 \, \text{m/sec} \)
After 7 seconds, the position will be:
- \( x = v_x \cdot t \approx 0.059 \times 7 \approx 0.413 \, \text{m} \)
- \( y = v_y \cdot t \approx 1.497 \times 7 \approx 10.48 \, \text{m} \)
The closest answer available is:
**\( (0.5, 10.5) \)**
### 3. Time for the Slingshot Ball to Hit the Ground
The ball is thrown at 110 ft/sec at an angle of 20° from a height of 4.5 ft.
Vertical motion is affected by gravity:
- The vertical velocity \( v_y = 110 \sin(20°) \)
- The initial height \( h = 4.5 ft \) and acceleration \( a = -32 \, \text{ft/sec}^2 \)
The vertical position equation is:
\[
y(t) = h + v_y t - \frac{1}{2} g t^2
\]
Setting \( y(t) = 0 \) gives:
\[
0 = 4.5 + (110 \sin(20°)) t - \frac{32}{2} t^2
\]
Calculating \( v_y \):
- \( v_y = 110 \sin(20°) \approx 110 \cdot 0.342 \approx 37.62 \)
The formula becomes:
\[
0 = 4.5 + 37.62 t - 16 t^2
\]
Using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-37.62 \pm \sqrt{37.62^2 - 4 \cdot (-16) \cdot 4.5}}{2 \cdot -16}
\]
Calculating the terms gives approximately:
After solving this, we find:
- \( t \approx 6.46 \, \text{seconds} \)
So, the answer is:
**6.46 seconds**
All conclusions are based on standard physical equations and trigonometric identities.
### 1. Polar Form of Parametric Equations
The parametric equations given are:
- \( x = 4t \)
- \( y = t^2 \)
First we need to express \( y \) in terms of \( x \):
- From \( x = 4t \), we can solve for \( t \):
\[
t = \frac{x}{4}
\]
Now substitute this expression for \( t \) into the equation for \( y \):
- \( y = t^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \)
We also have the relation between Cartesian coordinates and polar coordinates:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
Since \( y = \frac{x^2}{16} \), we can express it in terms of \( r \) and \( \theta \):
- \( r \sin \theta = \frac{(r \cos \theta)^2}{16} \)
- \( r \sin \theta = \frac{r^2 \cos^2 \theta}{16} \)
Assuming \( r \neq 0 \), we can divide both sides by \( r \):
- \( \sin \theta = \frac{r \cos^2 \theta}{16} \)
- Rearranging gives us:
\[
r = 16 \frac{\sin \theta}{\cos^2 \theta}
= 16 \tan \theta \sec \theta
\]
So the answer to the first part is:
**\( r = 16 \tan \theta \sec \theta \)**
### 2. Position of the Bowling Ball
The bowling ball is thrown with a velocity of 1.5 m/sec at an angle of 87°:
- The horizontal and vertical components of the velocity can be calculated as follows:
- \( v_x = v \cos(\theta) = 1.5 \cos(87°) \)
- \( v_y = v \sin(\theta) = 1.5 \sin(87°) \)
Using \( \cos(87°) \approx 0.0393 \) and \( \sin(87°) \approx 0.998 \):
- \( v_x \approx 1.5 \times 0.0393 \approx 0.059 \, \text{m/sec} \)
- \( v_y \approx 1.5 \times 0.998 \approx 1.497 \, \text{m/sec} \)
After 7 seconds, the position will be:
- \( x = v_x \cdot t \approx 0.059 \times 7 \approx 0.413 \, \text{m} \)
- \( y = v_y \cdot t \approx 1.497 \times 7 \approx 10.48 \, \text{m} \)
The closest answer available is:
**\( (0.5, 10.5) \)**
### 3. Time for the Slingshot Ball to Hit the Ground
The ball is thrown at 110 ft/sec at an angle of 20° from a height of 4.5 ft.
Vertical motion is affected by gravity:
- The vertical velocity \( v_y = 110 \sin(20°) \)
- The initial height \( h = 4.5 ft \) and acceleration \( a = -32 \, \text{ft/sec}^2 \)
The vertical position equation is:
\[
y(t) = h + v_y t - \frac{1}{2} g t^2
\]
Setting \( y(t) = 0 \) gives:
\[
0 = 4.5 + (110 \sin(20°)) t - \frac{32}{2} t^2
\]
Calculating \( v_y \):
- \( v_y = 110 \sin(20°) \approx 110 \cdot 0.342 \approx 37.62 \)
The formula becomes:
\[
0 = 4.5 + 37.62 t - 16 t^2
\]
Using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-37.62 \pm \sqrt{37.62^2 - 4 \cdot (-16) \cdot 4.5}}{2 \cdot -16}
\]
Calculating the terms gives approximately:
After solving this, we find:
- \( t \approx 6.46 \, \text{seconds} \)
So, the answer is:
**6.46 seconds**
All conclusions are based on standard physical equations and trigonometric identities.