To find the parametric form of the tangent line to the image of f(t) at t=-4, we need to find the derivative of f(t) and evaluate it at t=-4.
The derivative of f(t) = ⟨3t^2, 5t, t-4⟩ with respect to t is f'(t) = ⟨6t, 5, 1⟩.
Evaluating f'(t) at t=-4, we get f'(-4) = ⟨6(-4), 5, 1⟩ = ⟨-24, 5, 1⟩.
Now, let's find a point on the image of f(t) at t=-4. Plugging in t=-4 into f(t), we get f(-4) = ⟨3(-4)^2, 5(-4), (-4)-4⟩ = ⟨48, -20, -8⟩.
So, a point on the image of f(t) at t=-4 is ⟨48, -20, -8⟩.
Now, we can use the point ⟨48, -20, -8⟩ and the direction vector ⟨-24, 5, 1⟩ of the tangent line to write the parametric form of the line.
L(t) = ⟨48, -20, -8⟩ + t⟨-24, 5, 1⟩
= ⟨48-24t, -20+5t, -8+t⟩
Therefore, the parametric form of the tangent line to the image of f(t) at t=-4 is L(t) = ⟨48-24t, -20+5t, -8+t⟩.
Find the parametric form of the tangent line to the image of f(t)=⟨3t2,5t,t−4⟩ at t=−4
L(t) =
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