Question

Find an equation of the tangent to the curve at the given point. x = cos(3θ) + sin(2θ), y = sin(3θ) + cos(2θ); θ = 0

Answer 1

The equation of the tangent to the curve at θ = 0 is y = (3x + 1)/2

To find the equation of the tangent, we need to find dy/dx ,the tangent to the curve.

So, dy/dx = dy/dθ ÷ dx/dθ

Finding dy/dθ

Since y = sin(3θ) + cos(2θ)

dy/dθ = d[sin(3θ) + cos(2θ)]/dθ

dy/dθ = dsin(3θ)/dθ + dcos(2θ)/dθ

dy/dθ = 3cos(3θ) + [-2sin(2θ)]

dy/dθ = 3cos(3θ) - 2sin(2θ)

Finding dx/dθ

Also, x = cos(3θ) + sin(2θ)

dx/dθ = d[cos(3θ) + sin(2θ)]/dθ

dx/dθ = dcos(3θ)/dθ + dsin(2θ)/dθ

dx/dθ = -3sin(3θ) + 2cos(2θ)

dx/dθ = 2cos(2θ) - 3sin(3θ)

Finding dy/dx

So, dy/dx = dy/dθ ÷ dx/dθ

dy/dx = 3cos(3θ) - 2sin(2θ) ÷ [2cos(2θ) - 3sin(3θ)]

when θ = 0

dy/dx = 3cos(3 × 0) - 2sin(2 × 0) ÷ [2cos(2 × 0) - 3sin(3 × 0)]

dy/dx = 3cos(0) - 2sin(0) ÷ [2cos(0) - 3sin(0)]

dy/dx = 3 × 1 - 2 × 0 ÷ [2 × 1 - 3 × 0]

dy/dx = 3 - 0 ÷ [2 - 0]

dy/dx = 3/2

We now find the values of y and x when θ = 0.

Finding the value of y

So, y = sin(3θ) + cos(2θ)

y = sin(3 × 0) + cos(2 × 0)

y = sin(0) + cos(0)

y = 0 + 1

y = 1

Finding the value of x

Also, x = cos(3θ) + sin(2θ)

x = cos(3 × 0) + sin(2 × 0)

x = cos(0) + sin(0)

x = 1 + 0

x = 1

Finding the equation of the tangent

We now find the tangent to the curve from (y - y₀)/(x - x₀) = dy/dx

So, with x₀ = 1, y₀ = 1 and dy/dx = 3/2, we have

(y - y₀)/(x - x₀) = dy/dx

(y - 1)/(x - 1) = 3/2

y - 1 = 3/2(x - 1)

y = 3x/2 - 1/2 + 1

y = 3x/2 + 1/2

y = (3x + 1)/2

So, the equation of the tangent to the curve at θ = 0 is y = (3x + 1)/2

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