Question

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter: x = t³ - 1, y = t¹⁰ - t; t = -1.

Answer 1

To find the equation of the tangent to the curve at the given point, we need to find the derivative of the curve and substitute the coordinates of the point into the derivatives to find the slope of the tangent. Then, we can use the point-slope form of a linear equation to find the equation of the tangent.

Step-by-step explanation:

To find the equation of the tangent to the curve at the given point, we first need to find the derivative of the curve. The derivative represents the slope of the curve at any given point.

Let's differentiate the given curve with respect to t:

x' = 3t^2

y' = 10t^9 - 1

Now substitute t = -1 into the derivatives to find the slope of the tangent at this point:

x'(-1) = 3(-1)^2 = 3

y'(-1) = 10(-1)^9 - 1 = -11

So, the slope of the tangent is -11/3. Now we can use the point-slope form of a linear equation to find the equation of the tangent:

y - y1 = m(x - x1)

Substituting the coordinates of the given point (x1, y1) = (-1, -1) and the slope m = -11/3:

y + 1 = (-11/3)(x + 1)

Simplifying the equation:

3y + 3 = -11x - 11

Now we can rearrange the equation to get the standard form of a linear equation:

11x + 3y = -14

Therefore, the equation of the tangent to the curve at the point corresponding to t = -1 is 11x + 3y = -14.