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kkurt [141]
3 years ago
14

Find the solutions in the interval [0, 2π). (Enter your answers as a comma-separated list.)sec θ − tan θ = cos θ

Mathematics
1 answer:
AURORKA [14]3 years ago
5 0

Answer:

<h3>π/2</h3>

Step-by-step explanation:

Given the expression;

secθ − tan θ = cosθ

We are to find the solution in the interval [0, 2π).This is as shown;

From trigonometry identity;

secθ = 1/cosθ

tanθ = sinθ/cosθ

Substitute into the formula;

secθ − tan θ = cosθ

1/cosθ-sinθ/cosθ = cosθ

Multiply through by cosθ

1 - sinθ = cos²θ

1-sinθ = (1-sin²θ)

1-sin²θ-1+sinθ =0

-sin²θ+sinθ = 0

sin²θ = sinθ

sinθ = 1

θ = arcsin 1

θ = 90

θ = π/2

Hence the solution is π/2

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Compute the second partial derivatives ∂2f ∂x2 , ∂2f ∂x ∂y , ∂2f ∂y ∂x , ∂2f ∂y2 for the following function. f(x, y) = 2xy (x2 +
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Answer with step-by-step explanation:

We are given that a function

f(x,y)=2xy(x^2+y^2)^2

Differentiate partially w.r.t x

Then, we get

\frac{\delta f}{\delta x}=2y(x^2+y^2)^2+8x^2y(x^2+y^2)=(x^2+y^2)(2x^2y+2y^3+8x^2y)=2(5x^2y+y^3)(x^2+y^2)

Differentiate again w.r.t x

\frac{\delta^2f}{\delta x^2}=2(10xy)(x^2+y^2)+4x(5x^2y+y^3)=20x^3y+20xy^3+20x^3y+4xy^3=40x^3y+24xy^3

Differentiate function w.r.t y

\frac{\delta f}{\delta y}=2x(x^2+y^2)^2+2xy\times 2(x^2+y^2)\times 2y

\frac{\delta f}{\delta y}=(x^2+y^2)(2x^3+2xy^2+8xy^2)=2(x^2+y^2)(x^3+5xy^2)

Again differentiate w.r.t y

\frac{\delta^2f}{\delta x^2}=2(2y)(x^3+5xy^2)+20xy(x^2+y^2)=4x^3y+20xy^3+20x^3y+20xy^3=24x^3y+40xy^3

Differentiate partially w.r.t y

\frac{\delta^2f}{\delta y\delta x}=2(2y(5x^2y+y^3)+(x^2+y^2)(5x^2+3y^2))=10x^4+36x^2y^2+10y^4

\frac{\delta^2f}{\delta y\delta x}=10x^4+36x^2y^2+10y^4\frac{\delta^2f}{\delta x\delat y}=2(2x(x^3+5xy^2)+(3x^2+5y^2)(x^2+y^2))=10x^4+36x^2y^2+10y^4

\frac{\delta^2f}{\delta x\delat y}=10x^4+36x^2y^2+10y^4

Hence, if f(x,y) is of class C^2 (is twice continuously differentiable), then the mixed partial derivatives are equal.

i.e\frac{\delta^2f}{\delta y\delta x}=\frac{\delta^2f}{\delta x\delta y}

8 0
4 years ago
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