Find an equation for the line tangent to the curve at the...

Find an equation for the line tangent to the curve at
the point defined by the given value of $t$ . Also, find the value of $d^{2} y / d x^{2}$
at this point.
$$
x=\cos t, \quad y=1+\sin t, \quad t=\pi / 2
$$

Key Concepts

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter. This framework allows the representation of a curve in the plane without explicitly solving one variable in terms of another, making it easier to analyze curves and their properties using calculus.

First Derivative (dy/dx) in Parametric Form

For a curve defined by parametric equations, the first derivative dy/dx is found using the formula (dy/dt)/(dx/dt). This derivative represents the slope of the tangent line to the curve at a given point and is essential for understanding the instantaneous rate of change along the curve.

Tangent Line Equation

The equation of the tangent line to a curve at a specific point is determined by using the point-slope form of a line, with the coordinates of the point and the slope obtained from the first derivative (dy/dx). This line approximates the curve near the point of tangency.

Second Derivative (d²y/dx²) in Parametric Form

The second derivative for a parametric curve is computed by taking the derivative of the first derivative with respect to the parameter and then dividing by dx/dt. This measure describes the curvature of the curve, indicating how the slope of the tangent line changes along the curve.

Transcript

00:01 Let's consider the parametric equations defined by x equal to cosine of t and y equal to 1 plus sine of t.

00:06 Now we want to find the equation of the tangent line to the function described by these parametric equations at the point t equals pi over 2.

00:17 Now we recall that the equation of the tangent line is given by y equal to m times x minus x sub 0 plus y sub 0, wherein m is just the slope of the tangent line defined by dy over dx, but since this is with respect to t, we have to evaluate it at t which equals pi over 2.

00:41 And our x sub 0 here is going to be x evaluated at pi over 2, while y sub 0 is y evaluated at pi over 2.

00:52 So first you want to find dy over dx.

00:55 Dy over dx is defined to be the quotient of dy over dt and dx over dt.

01:06 Based on what we have, dy over dt are the derivative with respect to t of y, so it would be the derivative with respect to t of 1 plus sine of t, and dx over dt is the derivative with respect to of cosine of t.

01:23 And this will just give us cosine of t over negative sine of t, or that's negative cotangent of t.

01:33 And then from there we will find the value of the slope m, so at t which equals pi over 2, we have m equal to dy over dx evaluated at pi over 2, and that's going to be cosine of pi over 2 over negative sine of pi over 2, which is 0 over negative 1, which equals 0.

02:02 And then for x sub 0, it's going to be x evaluated at pi over 2, it's cosine of pi over 2 which is 0, and y sub 0 is going to be y evaluated at pi over 2, which in this case will be 1 plus sine of pi over 2, and that's 1 plus 1 equal to 2...

Please give Ace some feedback