Question

Find an equation tor the line tangent to the curve at the point defined by the given value of A. Also, find the value of de? d 2 y ​ at this point x=sec 2 1−1.y=cos1;1=− 3 π ​ Write the equation of the tangent line (Typo exact answerti, using radicalts as needed)

Answer 1

The equation of the tangent line to the curve y = cos(1) at the point where x = sec^2(1) is y = -3π(x - sec^2(1)) + cos(1). Additionally, the value of d^2y/dx^2 at this point is 6π.

Step-by-step explanation:

To find the equation of the tangent line, we begin by determining the slope of the tangent line at the given point. The derivative dy/dx of the function y = cos(1) is obtained as -sin(1). Substituting x = sec^2(1) into this derivative gives the slope of the tangent line at that specific point. Now, using the point-slope form of a line, y - y₁ = m(x - x₁), where (x₁, y₁) is the point of tangency and m is the slope, we substitute in the values to obtain the equation of the tangent line.

In this case, the point of tangency is (sec^2(1), cos(1)), and the slope is -sin(1) evaluated at x = sec^2(1). Therefore, the equation of the tangent line is y = -sin(1)(x - sec^2(1)) + cos(1). To simplify this expression, we can use the trigonometric identity sin^2(1) + cos^2(1) = 1, which yields sin^2(1) = 1 - cos^2(1). Substituting this into the equation of the tangent line, we get y = -√(1 - cos^2(1))(x - sec^2(1)) + cos(1). Further simplification leads to the final answer provided.

Now, for the second part of the question regarding the value of d^2y/dx^2, we take the second derivative of the original function y = cos(1). The second derivative, d^2y/dx^2, is equal to -cos(1). Substituting x = sec^2(1) into this expression gives the value of the second derivative at the specified point, which is -cos(1) = -3π.

Answer 2

1. Equation of the tangent line:
`\(y = -√(2)(x - \sec^2(1)) + \cos(1)\)`

2.
`\((d^2y)/(dx^2)\) at \(x = \sec^2(1)\): \((2√(2))/(\cos^3(1))\)`

Step-by-step explanation:

The equation of the tangent line is found using the point-slope form
`\(y - y_1 = m(x - x_1)\)`, where
`\(m\)` is the slope and
`\((x_1, y_1)\)` is the point of tangency. The slope
`\(m\)` is determined by finding the first derivative
`\((dy)/(dx)\)` and evaluating it at the given
`\(x\)`-value. Then, the tangent line equation is obtained by substituting the slope and point of tangency into the point-slope form.

For this specific problem, the slope
`\(m\) is \(-√(2)\)` after finding
`\((dy)/(dx)\)` and evaluating it at
`\(x = \sec^2(1)\)`. The point of tangency. Substituting these values into the point-slope form yields the equation of the tangent line.

To find
`\((d^2y)/(dx^2)\) at \(x = \sec^2(1)\)`, the second derivative is computed and then evaluated at the given
`\(x\)`-value. In this case,
`\((d^2y)/(dx^2) = (2√(2))/(\cos^3(1))\)`. This value represents the concavity of the curve at the point of tangency, indicating whether the curve is concave up or down at that specific
`\(x\)`-value.