Answer:
k(t) = (sin3, cos3, 2) + [(t - 1)(cos3, -3sin3, 7)]
Step-by-step explanation:
For a path r(t), the general equation k(t) of its tangent line at a specified point r(t₀) is given by;
k(t) = r(t₀) + r'(t₀) [t - t₀] -----------------(i)
Where
r'(t) is the first derivative of the path r(t) at a given value of t.
From the question:
r(t) = (sin3t, cos3t, 2[tex]t^{7/2}[/tex]) and t₀ = 1
=> r(1) = (sin3, cos3, 2) at t₀ = 1
Find the first derivative component-wise of r(t) to get r'(t)
∴ r'(t) = (cos3t, -3sin3t, 7[tex]t^{5/2}[/tex])
=>r'(1) = (cos3, -3sin3, 7)
Now, at t₀ = 1, equation (i) becomes;
k(t) = r(1) + [ r'(1) (t-1)] [substitute the necessary values]
k(t) = (sin3, cos3, 2) + [(t - 1)(cos3, -3sin3, 7)]
