find the tangent line of (t^3 -2t+1)/3t at t=2

To find the equation of the tangent line of the function

\[
f(t) = \frac{t^3 - 2t + 1}{3t}
\]

at \( t = 2 \), we will follow these steps:

### Step 1: Differentiate the function to find \( f'(t) \)
Using the quotient rule, which states:

\[
\left( \frac{u(t)}{v(t)} \right)' = \frac{u'(t)v(t) - u(t)v'(t)}{v(t)^2}
\]

where \( u(t) = t^3 - 2t + 1 \) and \( v(t) = 3t \), we need to find \( u'(t) \) and \( v'(t) \).

1. \( u'(t) = 3t^2 - 2 \)
2. \( v'(t) = 3 \)

Now applying the quotient rule:

\[
f'(t) = \frac{(3t^2 - 2)(3t) - (t^3 - 2t + 1)(3)}{(3t)^2}
\]

### Step 2: Simplify the derivative
First, simplify the numerator:

\[
(3t^2 - 2)(3t) = 9t^3 - 6t
\]

\[
(t^3 - 2t + 1)(3) = 3t^3 - 6t + 3
\]

Now subtract these:

\[
(9t^3 - 6t) - (3t^3 - 6t + 3) = 9t^3 - 6t - 3t^3 + 6t - 3 = 6t^3 - 3
\]

So the derivative becomes:

\[
f'(t) = \frac{6t^3 - 3}{9t^2}
\]

Simplifying further:

\[
f'(t) = \frac{2t^3 - 1}{3t^2}
\]

### Step 3: Evaluate \( f'(t) \) at \( t = 2 \)
Now, substitute \( t = 2 \) into the derivative:

\[
f'(2) = \frac{2(2)^3 - 1}{3(2)^2} = \frac{2(8) - 1}{3(4)} = \frac{16 - 1}{12} = \frac{15}{12} = \frac{5}{4}
\]

So the slope of the tangent line at \( t = 2 \) is \( \frac{5}{4} \).

### Step 4: Find \( f(2) \)
Now, substitute \( t = 2 \) into the original function to find \( f(2) \):

\[
f(2) = \frac{(2)^3 - 2(2) + 1}{3(2)} = \frac{8 - 4 + 1}{6} = \frac{5}{6}
\]

### Step 5: Use the point-slope form to find the equation of the tangent line
The point-slope form of the equation of a line is:

\[
y - y_1 = m(t - t_1)
\]

where \( m = \frac{5}{4} \), \( t_1 = 2 \), and \( y_1 = f(2) = \frac{5}{6} \). Plugging these values into the equation:

\[
y - \frac{5}{6} = \frac{5}{4}(t - 2)
\]

Simplifying this:

\[
y = \frac{5}{4}(t - 2) + \frac{5}{6}
\]

You can leave the equation in this form or simplify further as desired.

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Would you like more details on any specific part of this process?

### Related questions:
1. What is the geometrical interpretation of the derivative?
2. How can we simplify the tangent line equation further?
3. How does the quotient rule apply to other functions?
4. What happens to the tangent line if we change the point \( t = 2 \)?
5. How do we find the second derivative of this function?

#### Tip:
When using the quotient rule, it’s helpful to simplify each component as much as possible before applying the rule to avoid complex algebraic steps later.

Find the equation of the tangent line of the function (t^3 - 2t + 1)/3t at t = 2 using the quotient rule and differentiation. This calculus problem demonstrates how to compute the slope of the tangent line and use point-slope form to derive the equation. Suitable for Grades 11-12.

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