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# Determine the equation of the tangent line to the given path at the specified value of t. (sin 3t, cos3t, 2t^{5/2}); 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(t_{0}) + f'(t_{0})(t - t_{0})

f(t_{0}) = (sin(3t), cos(3t), 2t^{5/2})----->(1)

When t_{0} = 1

f(t_{0}) = {sin3(1), cos3(1), 2(1)^{5/2}}

f(t_{0}) = {sin3, cos3, 2}

Now differentiate (1) with respect to t

f'(t_{0}) = (3cos3t, -3sin3t, 5/2{2t^{5/2-1}})

f'(t_{0}) = (3cos3t, -3sin3t, 5t^{3/2}}

When t_{0} = 1

f'(1) = (3cos3(1), -3sin3(1), 5(1)^{3/2})

f'(1) =(3cos3, -3sin3, 5)

By substituting f(t_{0}) and f'(t_{0}) in the given function,

P(t) = f(t_{0}) + f'(t_{0})(t - t_{0})

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, 2t^{5/2}); t = 1.

**Summary:**

The equation of the tangent line to the given path at the specified value of t in (sin(3t), cos(3t), 2t^{5/2}); t = 1 is P(t) = {sin3, cos3, 2} + (3cos3, -3sin3, 5)(t - 1).

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