To find the equation of the tangent line to the function $f(t) = 3t^2 \cos(t)$ at $t = \pi$, we need to follow these steps:

1. Calculate $f(\pi)$ to find the point where the tangent touches the curve.
2. Differentiate $f(t)$ with respect to $t$ to get $f'(t)$, the slope of the tangent line.
3. Evaluate $f'(\pi)$ to find the slope at $t = \pi$.
4. Use the point-slope form of the tangent line: $y - f(\pi) = f'(\pi) (t - \pi)$.

Step 1: Calculate $f(\pi)$

$f(\pi) = 3\pi^2 \cos(\pi) = 3\pi^2 \cdot (-1) = -3\pi^2$ So, the point on the curve at $t = \pi$ is $(\pi, -3\pi^2)$.

Step 2: Differentiate $f(t)$ to find $f'(t)$

Since $f(t) = 3t^2 \cos(t)$, we apply the product rule: $f'(t) = \frac{d}{dt}(3t^2) \cdot \cos(t) + 3t^2 \cdot \frac{d}{dt}(\cos(t))$ $= (6t) \cos(t) + 3t^2 \cdot (-\sin(t))$ $= 6t \cos(t) - 3t^2 \sin(t)$

Step 3: Evaluate $f'(\pi)$

Substitute $t = \pi$ into $f'(t)$: $f'(\pi) = 6\pi \cos(\pi) - 3\pi^2 \sin(\pi)$ Since $\cos(\pi) = -1$ and $\sin(\pi) = 0$: $f'(\pi) = 6\pi \cdot (-1) - 3\pi^2 \cdot 0 = -6\pi$ The slope of the tangent line at $t = \pi$ is $-6\pi$.

Step 4: Write the equation of the tangent line

Using the point-slope form $y - y_1 = m(t - t_1)$, where $m = -6\pi$, $t_1 = \pi$, and $y_1 = -3\pi^2$: $y + 3\pi^2 = -6\pi (t - \pi)$ Simplifying, $y = -6\pi t + 6\pi^2 - 3\pi^2$ $y = -6\pi t + 3\pi^2$

Final Answer

The equation of the tangent line to $f(t) = 3t^2 \cos(t)$ at $t = \pi$ is: $y = -6\pi t + 3\pi^2$

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Related Questions

1. How do we find the tangent line of $f(t) = t^3 \sin(t)$ at $t = \pi/2$?
2. What is the derivative of $g(t) = e^t \cos(t)$ using the product rule?
3. How do we find the tangent line of a trigonometric function at a specific point?
4. What is the geometric interpretation of the slope of the tangent line?
5. How does the product rule apply when differentiating $h(t) = t^2 \sin(t^2)$?

Tip: Remember that the product rule states: if $u(t)$ and $v(t)$ are functions of $t$, then $\frac{d}{dt} [u(t)v(t)] = u'(t)v(t) + u(t)v'(t)$.