Find the parametric form of the tangent line to the image of f(t)=⟨3t2,5t,t−4⟩ at t=−4
L(t) =
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⟩.
To find the parametric form of the tangent line to the image of f(t) = ⟨3t2, 5t, t−4⟩ at t = −4, we first need to find the derivative of the function at t = −4.
Step 1: Find the derivative of f(t):
f'(t) = ⟨6t, 5, 1⟩
Step 2: Evaluate f'(t) at t = −4:
f'(-4) = ⟨6(-4), 5, 1⟩ = ⟨-24, 5, 1⟩
Step 3: Use the derivative and the given point to write the equation of the tangent line:
The tangent line will pass through the point f(-4) = ⟨3(-4)2, 5(-4), (-4)−4⟩ = ⟨48, -20, -8⟩
Let's use the point-slope form of a line equation: (x - x₁)/a = (y - y₁)/b = (z - z₁)/c
Substituting the values:
(x - 48)/-24 = (y - (-20))/5 = (z - (-8))/1
Simplifying, we get:
(x - 48)/-24 = (y + 20)/5 = (z + 8)/1
Rearranging, we can write the parametric form of the tangent line as follows:
x = -24t + 48
y = 5t - 20
z = t + 8
Therefore, the parametric form of the tangent line to the image of f(t) = ⟨3t2, 5t, t−4⟩ at t = −4 is:
x = -24t + 48
y = 5t - 20
z = t + 8