# Determine the equation of the tangent line to the given path at the specified value of t. (sin(5t), cos(5t), 2t^{7/2}); t = 1

**Solution:**

The given problem statement is the problem of Linearization, which is based on tangent lines. Linearization concept is based on the principle that when we draw a tangent to a curve at a point to the curve the tangent line gives good approximations to the y-values on the curve. Locally every differential curve behaves like a straight line.

So if f is differentiable at t = a then the approximating function

L(t) = f(a) + f’(a)(t - a) is the linearization of f at a.

The approximation f(t) ≈ L(t) of f by L is the standard linear approximation of f at a. The point x = a is the center of the approximation.

a) f(t) = sin5t

f(t = a = 1) = sin5

Differential of f(t) is given as

f’(t) = 5cos5t

f’(t = a = 1) = 5cos5

Therefore

L(t) = f(1) + f’(1) × (t -1)

= sin5 + 5cos5 × (t - 1)

= sin5 + 5tcos5 - 5cos5

= 0.0871+ 5t(0.9961) - 5(0.9961)

= 0.0871 + 4.9805t - 4.9805

**L(t) = -4.8934+ 4.9805t**

b) f(t) = cos5t

f(t = a = 1) = cos5 = 0.9961

f’(t) = -5sin5t

f’(t = a = 1) = 5sin5 = -5 × (0.0871) = -0.4355

Therefore

L(t) = f(1) + f’(1) × (t - 1)

= 0.9961 + -0.4355(t - 1)

=0.9961 + -0.4355t + 0.4355

**L(t) =1.4316 - 0.4355t**

c) f(t) = 2t^{7/2}

f(t = a = 1) = 2(1)^{7/2} = 2

f’(t) = 2 × (7/2)t^{5/2}

= 7t^{5/2}

f’(t = a =1) = 7(1)^{5/2} = 7

L(t) = f(1) + f’(1) × (t -1)

= 2 + 7(t - 1)

= 2 + 7t - 7

**L(t) = 7t - 5**

## Determine the equation of the tangent line to the given path at the specified value of t. (sin(5t), cos(5t), 2t^{7/2}); t = 1

**Summary:**

The equations of the tangent line to the given path at the specified value of t for sin(5t), cos(5t) and 2t^{7/2}); t = 1 are 4.9805t - 4.8934, 1.4316 - 0.4355t and 7t - 5 respectively.

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