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Magazine Chapter 11: Problem 48
Find the tangential and normal components \(\left(a_{T}\right.\) and \(\left.a_{N}\right)\) of the acceleration vector at \(t .\) Then evaluate at \(t=t_{1}\). $$ x=t, y=t^{2}, z=t^{3} ; t_{1}=2 $$
Short Answer
Expert verified
At \( t = 2 \), the tangential component \( a_T \approx 14.49 \) and normal component \( a_N \approx 0.55 \).
Step by step solution
01
Find the velocity vector
The velocity vector \( \mathbf{v}(t) \) can be found by differentiating the position vector \( \mathbf{r}(t) = (t, t^2, t^3) \) with respect to time \( t \). Differentiate each component to get: \( \mathbf{v}(t) = (1, 2t, 3t^2) \).
02
Find the acceleration vector
The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) = (1, 2t, 3t^2) \). Differentiate each component to obtain: \( \mathbf{a}(t) = (0, 2, 6t) \).
03
Compute the magnitude of the velocity vector
To find the tangential component, first compute the magnitude of the velocity vector: \( \| \mathbf{v}(t) \| = \sqrt{1^2 + (2t)^2 + (3t^2)^2} = \sqrt{1 + 4t^2 + 9t^4} \). Evaluate this at \( t = 2 \) later.
04
Tangential component of the acceleration
The tangential component of the acceleration, \( a_T \), is given by the formula \( a_T = \frac{\mathbf{v}(t) \cdot \mathbf{a}(t)}{\| \mathbf{v}(t) \|} \). Calculate the dot product \( \mathbf{v}(t) \cdot \mathbf{a}(t) = 1 \times 0 + 2t \times 2 + 3t^2 \times 6t = 4t + 18t^3 \). So, \( a_T = \frac{4t + 18t^3}{\| \mathbf{v}(t) \|} \).
05
Normal component of the acceleration
The normal component of the acceleration, \( a_N \), can be found using the formula \( a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \). First, find \( \| \mathbf{a}(t) \| = \sqrt{0^2 + 2^2 + (6t)^2} = \sqrt{4 + 36t^2} = \sqrt{4 + 36t^2} \).
06
Evaluate at \( t = t_1 = 2 \)
Evaluate each expression at \( t = 2 \). Compute \( \| \mathbf{v}(2) \| = \sqrt{1 + 16 + 144} = \sqrt{161} \), \( \mathbf{v}(2) \cdot \mathbf{a}(2) = 40 + 144 = 184 \), and \( a_T = \frac{184}{\sqrt{161}} \). Compute \( \| \mathbf{a}(2) \| = \sqrt{148} \) and finally \( a_N = \sqrt{148 - \left(\frac{184}{\sqrt{161}}\right)^2} \).
07
Calculate final components
Finally, simply calculate the numerical expressions for the components: \( a_T = \frac{184}{\sqrt{161}} \approx 14.49 \) and \( a_N = \sqrt{148 - \left(\frac{184}{\sqrt{161}}\right)^2} \approx 0.55 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tangential Component
The tangential component of acceleration, denoted as \(a_T\), reflects how much the object's speed changes along its path. It's like the gas pedal in a car, influencing how fast you're moving forwards or backwards at any moment.
To compute \(a_T\), you need to take the dot product of the velocity vector, \( \mathbf{v}(t) \), and the acceleration vector, \( \mathbf{a}(t) \). This dot product gives a single number that tells us how much of the acceleration points in the direction of movement.
To compute \(a_T\), you need to take the dot product of the velocity vector, \( \mathbf{v}(t) \), and the acceleration vector, \( \mathbf{a}(t) \). This dot product gives a single number that tells us how much of the acceleration points in the direction of movement.
- First, differentiate the position vector \( \mathbf{r}(t) \) to get \( \mathbf{v}(t) = (1, 2t, 3t^2) \).
- Next, differentiate \( \mathbf{v}(t) \) to obtain \( \mathbf{a}(t) = (0, 2, 6t) \).
- The dot product \( \mathbf{v}(t) \cdot \mathbf{a}(t) \) is calculated to produce \( 4t + 18t^3 \).
- The magnitude of the velocity vector \( \| \mathbf{v}(t) \| \) becomes the denominator.
Normal Component
The normal component of acceleration, \( a_N \), is like the steering wheel in your car. It tells us how much the path is curving, even if the speed remains constant.
When calculating \( a_N \), you start by looking at the acceleration that isn't aligned with the direction of movement. This involves separating the total acceleration into the part...
When calculating \( a_N \), you start by looking at the acceleration that isn't aligned with the direction of movement. This involves separating the total acceleration into the part...
- First, calculate \( \| \mathbf{a}(t) \| \), which measures the overall magnitude of acceleration. Here, \( \| \mathbf{a}(t) \| = \sqrt{4 + 36t^2} \).
- Next, use the formula \( a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \) to find the normal component.
Velocity Vector
The velocity vector, \( \mathbf{v}(t) \), is a key player in understanding both speed and direction. Think of it as a way to describe where you're moving, not just how fast.
To find \( \mathbf{v}(t) \), differentiate each component of the position vector \( \mathbf{r}(t) = (t, t^2, t^3) \) with respect to time. This involves straightforward calculations:
\[\mathbf{v}(t) = (1, 2t, 3t^2)\].
This gives:
To find \( \mathbf{v}(t) \), differentiate each component of the position vector \( \mathbf{r}(t) = (t, t^2, t^3) \) with respect to time. This involves straightforward calculations:
\[\mathbf{v}(t) = (1, 2t, 3t^2)\].
This gives:
- \( 1 \) in the x-direction tells us the object consistently moves along x.
- \( 2t \) in the y-direction means your pace in the y-direction speeds up as \(t\) increases.
- \( 3t^2 \) in the z-direction suggests acceleration along z grows quadratically.
Acceleration Vector
The acceleration vector, \( \mathbf{a}(t) \), reveals how velocity changes over time and direction. Acceleration is about changes鈥攈ow swiftly and sharply movement occurs.
We determine \( \mathbf{a}(t) \) by differentiating the velocity vector \( \mathbf{v}(t) = (1, 2t, 3t^2) \). This means differentiating each component individually:
\[\mathbf{a}(t) = (0, 2, 6t) \].
This breakdown means:
We determine \( \mathbf{a}(t) \) by differentiating the velocity vector \( \mathbf{v}(t) = (1, 2t, 3t^2) \). This means differentiating each component individually:
\[\mathbf{a}(t) = (0, 2, 6t) \].
This breakdown means:
- The \(0\) in the x-direction shows that velocity doesn't change along x. Constant speed there.
- The \(2\) in the y-direction translates to constant acceleration, an unyielding pull on the pace along y.
- The \(6t\) in the z-direction implies substantial and increasing force impacting z as time moves ahead.
