To find the parametric form of the tangent line to the image of f(t) at t=-4, we first need to find the derivative of f(t). The derivative will give us the rate of change of f(t) at any given point.
The derivative of f(t) with respect to t is given by:
f'(t) = ⟨6t, 5, 1⟩
Now, we can evaluate f'(-4) to find the slope of the tangent line at t=-4.
f'(-4) = ⟨6(-4), 5, 1⟩ = ⟨-24, 5, 1⟩
So, the slope of the tangent line at t=-4 is -24 in the x-direction, 5 in the y-direction, and 1 in the z-direction.
Now, let's find a point on the image of f(t) at t=-4. Plugging t=-4 into f(t), we get:
f(-4) = ⟨3(-4)2, 5(-4), -4-4⟩ = ⟨48, -20, -8⟩
Therefore, the point on the image of f(t) at t=-4 is ⟨48, -20, -8⟩.
Finally, we can use the point-slope form of a line to write the equation of the tangent line:
L(t) = ⟨48, -20, -8⟩ + t⟨-24, 5, 1⟩
So, 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⟩
The parametric form of the tangent line to the image of f(t)=⟨3t2,5t,t−4⟩
at t=−4
is
L(t)=
1 answer