Determine the equation of the tangent line to the given path at the specified value of t. (sin 3t, cos3t, 2t5/2); t = 1.
Solution:
We know that the equation of the tangent line to the given path at the specified value t is written as
P(t) = f(t0) + f'(t0)(t - t0)
f(t0) = (sin(3t), cos(3t), 2t5/2)----->(1)
When t0 = 1
f(t0) = {sin3(1), cos3(1), 2(1)5/2}
f(t0) = {sin3, cos3, 2}
Now differentiate (1) with respect to t
f'(t0) = (3cos3t, -3sin3t, 5/2{2t5/2-1})
f'(t0) = (3cos3t, -3sin3t, 5t3/2}
When t0 = 1
f'(1) = (3cos3(1), -3sin3(1), 5(1)3/2)
f'(1) =(3cos3, -3sin3, 5)
By substituting f(t0) and f'(t0) in the given function,
P(t) = f(t0) + f'(t0)(t - t0)
P(t) = {sin3, cos3, 2} + (3cos3, -3sin3, 5)(t - 1)
Therefore, the equation of the tangent line is P(t) = {sin3, cos3, 2} + (3cos3, -3sin3, 5)(t - 1).
Determine the equation of the tangent line to the given path at the specified value of t. (sin 3t, cos3t, 2t5/2); t = 1.
Summary:
The equation of the tangent line to the given path at the specified value of t in (sin(3t), cos(3t), 2t5/2); t = 1 is P(t) = {sin3, cos3, 2} + (3cos3, -3sin3, 5)(t - 1).
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