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.
$$
x=\frac{1}{t+1}, \quad y=\frac{t}{t-1}, \quad t=2
$$

Key Concepts

Parametric Equations

Parametric equations represent curves using a pair (or set) of functions where each coordinate is expressed in terms of an independent parameter. This allows for a more flexible representation of curves, especially when the relationship between x and y is complicated or not a function in the usual sense. Parametric forms are widely used in calculus to analyze the geometric properties of curves.

Parametric Differentiation

Parametric differentiation is a technique used to compute derivatives of curves represented by parametric equations. Instead of directly finding dy/dx, one differentiates both x and y with respect to the parameter and then finds dy/dx by dividing dy/dt by dx/dt. This method is essential when the curve's relationship between x and y is given in terms of another parameter.

Tangent Line Determination

The tangent line to a curve at a given point is the straight line that just touches the curve at that point without crossing it. In the context of parametric equations, after finding the derivative dy/dx, which gives the slope of the tangent line, the point-slope form of a linear equation can be used to express the tangent line’s equation. Calculating the tangent line is fundamental in differential calculus and analysis of curves.

Second Derivative in Parametric Form

The second derivative of a function, particularly d²y/dx², provides information about the curvature or concavity of the curve at a given point. For parametric equations, it is computed by differentiating the first derivative dy/dx with respect to the parameter and then dividing by dx/dt. This process employs the chain rule and gives insight into the behavior of the curve beyond its immediate slope.

Transcript

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

00:07 Now in here we want to find the equation of the tangent line to the curve defined by these parametric equations at t which equals 2.

00:17 So first we note that the equation of the tangent line is defined by y equal to m times x minus x sub 0 plus y sub 0, wherein m is just the derivative or the slope of the tangent line defined by the derivative dy over dx evaluated at the given value of t.

00:39 And x sub 0 here is going to be x evaluated at the given value of t and y sub 0 is going to be y evaluated at the value or the given value of t.

00:52 So first we want to get dy over dx and dy over dx is just the derivative of y with respect to t all over the derivative of x with respect to t.

01:06 So that equals d over dt of t over t minus 1 over d over dt of 1 over t plus 1.

01:18 So for the numerator that's t minus 1 times the derivative of t which is 1 minus t times the derivative of t minus 1 which is 1 over t minus 1 squared.

01:34 All over we have negative of t plus 1 raised to negative 2.

01:41 So this will simplify into negative 1 over t minus 1 squared times the negative of t plus 1 raised to the second power over 1.

01:55 And that will simplify further into t plus 1 squared over t minus 1 squared.

02:05 So at t which equals 2, value of m is going to be 2 plus 1 squared over 2 minus 1 squared that will be equal to 9.

02:22 And for the value of x sub 0 that will be 1 over 2 plus 1 or 1 third...

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