The curve y^(2)sin x-2x=9-pi passes through (pi//2,3). Use local linearization to estimate the value of y at x=1.67. Use 1.57 for pi//2. Round to 2 d.p.

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Find Derivative: First, we need to find the derivative of the given curve with respect to x. The curve is defined by the equation y^2sin(x) - 2x = 9 - pi. We will differentiate both sides of the equation with respect to x, treating y as a function of x (y=y(x)).Differentiate Equation: Differentiating y^2sin(x) with respect to x using the product rule and chain rule gives us 2y(dy/dx)sin(x) + y^2cos(x). Differentiating -2x with respect to x gives us -2. The right side of the equation, 9 - pi, is a constant, so its derivative is 0.Set Derivative Equal: Setting the derivative equal to zero, we get 2y(dy/dx)sin(x) + y^2cos(x) - 2 = 0. Now we need to solve for dy/dx.Solve for dy/dx: We plug in the point (pi/2, 3) into the derivative to find the slope of the tangent line at that point. Since sin(pi/2) = 1 and cos(pi/2) = 0, the equation simplifies to 2*3*(dy/dx)*1 + 3^2*0 - 2 = 0, which simplifies to 6(dy/dx) - 2 = 0.Substitute Point: Solving for dy/dx, we get dy/dx = 2/6 = 1/3. This is the slope of the tangent line at the point (pi/2, 3).Write Tangent Line: The equation of the tangent line at (pi/2, 3) can be written in point-slope form: y - 3 = (1/3)(x - pi/2). We will use this linear equation to estimate the value of y at x=1.67.Estimate y Value: We substitute x=1.67 and pi/2=1.57 into the tangent line equation to estimate y: y - 3 = (1/3)(1.67 - 1.57). This simplifies to y - 3 = (1/3)(0.1).Calculate Result: Calculating the right side of the equation gives us (1/3)(0.1) = 0.0333 (rounded to 4 decimal places for intermediate calculation).Final Answer: Adding 3 to both sides of the equation to solve for y gives us y = 3 + 0.0333, which equals 3.0333.Rounding the final answer to two decimal places, we get y ≈ 3.03.

The curve $y^{2} \sin x-2 x=9-\pi$ passes through $(\pi / 2,3)$ . Use local linearization to estimate the value of $y$ at $x=1.67$ . Use $1$ . $57$ for $\pi / 2$ . Round to $2$ d.p.

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The curve y2sin⁡x−2x=9−π y^{2} \sin x-2 x=9-\pi y2sinx−2x=9−π passes through (π/2,3) (\pi / 2,3) (π/2,3). Use local linearization to estimate the value of y y y at x=1.67 x=1.67 x=1.67. Use 111.575757 for π/2 \pi / 2 π/2. Round to 222 d.p.

The curve $y^{2} \sin x-2 x=9-\pi$ passes through $(\pi / 2,3)$ . Use local linearization to estimate the value of $y$ at $x=1.67$ . Use $1$ . $57$ for $\pi / 2$ . Round to $2$ d.p.