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allsm [11]
3 years ago
5

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x = cos(θ) + sin(10

θ) y = sin(θ) + cos(10θ) θ = 0 y(x) =
Mathematics
1 answer:
vivado [14]3 years ago
6 0

Answer:

Step-by-step explanation:

x = cos θ + sin(10θ)

y = sin θ + cos(10θ)

Take derivative with respect to θ:

dx/dθ = -sin θ + 10 cos(10θ)

dy/dθ = cos θ - 10 sin(10θ)

Divide:

dy/dx = (dy/dθ) / (dx/dθ)

dy/dx = (cos θ - 10 sin(10θ)) / (-sin θ + 10 cos(10θ))

Evaluate the derivative at θ=0:

dy/dx = (cos 0 - 10 sin 0) / (-sin 0 + 10 cos 0)

dy/dx = 1/10

Evaluate the parametric functions at θ=0:

x = cos 0 + sin 0 = 1

y = sin 0 + cos 0 = 1

Writing the equation of the tangent line in point-slope form:

y - 1 = 1/10 (x - 1)

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3 years ago
I need help with finding x. Can someone help me please?
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Step-by-step explanation:

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