The velocity of a particle moving along a straight line is
v
(
t
)

=
sec
(
t
)
tan
(
t
)
where v is measured in meters per second and t in seconds. Find the acceleration
a
(
t
)
.

What is the acceleration when
t
=

1
/
2
Give your answer accurate to
3
decimal places.

To find the acceleration \( a(t) \), we need to differentiate the velocity function \( v(t) \) with respect to time \( t \).

The given velocity function is:
\[ v(t) = \sec(t) \tan(t) \]

We need to find \( a(t) = \frac{dv(t)}{dt} \).

First, let's use the product rule for differentiation. The product rule states:
\[ \frac{d}{dt} [f(t)g(t)] = f'(t)g(t) + f(t)g'(t) \]

Let \( f(t) = \sec(t) \) and \( g(t) = \tan(t) \).

Then,
\[ f'(t) = \frac{d}{dt} [\sec(t)] = \sec(t) \tan(t) \]
\[ g'(t) = \frac{d}{dt} [\tan(t)] = \sec^2(t) \]

Applying the product rule:
\[ \frac{d}{dt} [\sec(t) \tan(t)] = \sec(t) \tan(t) \cdot \tan(t) + \sec(t) \cdot \sec^2(t) \]
\[ = \sec(t) \tan^2(t) + \sec^3(t) \]

Therefore, the acceleration function \( a(t) \) is:
\[ a(t) = \sec(t) \tan^2(t) + \sec^3(t) \]

Next, we need to find the acceleration when \( t = \frac{1}{2} \).

First, we calculate:
\[ \sec \left( \frac{1}{2} \right) = \frac{1}{\cos \left( \frac{1}{2} \right)} \]
\[ \tan \left( \frac{1}{2} \right) = \frac{\sin \left( \frac{1}{2} \right)}{\cos \left( \frac{1}{2} \right)} \]

Using these in the acceleration formula:
\[ a \left( \frac{1}{2} \right) = \sec \left( \frac{1}{2} \right) \left( \tan \left( \frac{1}{2} \right) \right)^2 + \left( \sec \left( \frac{1}{2} \right) \right)^3 \]

We can now plug in the values and simplify using numerical values:
\[ \sec \left( \frac{1}{2} \right) \approx 1.139 \]
\[ \tan \left( \frac{1}{2} \right) \approx 0.546 \]

Substituting these values in:
\[ a \left( \frac{1}{2} \right) \approx 1.139 \cdot (0.546)^2 + (1.139)^3 \]
\[ \approx 1.139 \cdot 0.298 + 1.476 \]
\[ \approx 0.339 + 1.476 \]
\[ \approx 1.815 \]

Therefore, the acceleration when \( t = \frac{1}{2} \) is approximately \( 1.815 \) meters per second squared.

Would you like more details or have any questions about the process?

Here are some questions you might ask next:
1. How do I differentiate more complex functions?
2. What are other applications of the product rule in calculus?
3. Can you explain the chain rule with examples?
4. How do I find the derivative of trigonometric functions?
5. What is the importance of understanding acceleration in physics?

**Tip:** Always remember to apply the correct differentiation rules such as the product rule, chain rule, or quotient rule based on the function you are working with.

Learn how to find acceleration from the velocity function v(t) = sec(t) tan(t) in calculus. Step-by-step solution provided, including differentiation using the product rule. Calculate acceleration at t = 1/2 accurate to 3 decimal places.

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