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.
$x=t, \quad y=\sqrt{t}, \quad t=1 / 4$.

Key Concepts

Parametric Equations

Parametric equations describe curves by defining each coordinate as a function of an independent parameter. This approach is useful for representing curves that cannot be easily described using a single function, and it plays a vital role in understanding complex trajectories in both two-dimensional and higher-dimensional spaces.

Tangent Line to a Parametric Curve

The tangent line to a parametric curve is the line that just touches the curve at a given point, possessing the same instantaneous direction as the curve. To find its equation, the slope obtained from the derivative dy/dx is used along with the coordinates of the point on the curve to establish a linear equation.

Second Derivative in Parametric Form

The second derivative, d²y/dx², provides information about the curvature of the parametric curve, such as concavity and points of inflection. It is calculated by differentiating the first derivative with respect to the parameter and then dividing by the derivative of the x-component, offering insights into the acceleration of change in the curve’s slope.

Differentiation of Parametric Equations

Differentiation in the context of parametric equations involves computing derivatives with respect to the parameter and then forming ratios to obtain dy/dx. This method helps determine the slope of the tangent at any specific point on the curve, which is crucial for analyzing the behavior and direction of the curve.

Transcript

00:01 This is a two -part question, and in the first part of the question, you are asked to find the equation of the tangent line when t equals one -fourth, if you're given that x of t equals t and y of t equals the square root of t.

00:20 All right, to get the equation of the tangent line, you need two things.

00:24 You need to know the slope, and we need to find out what the point is.

00:32 All right, so let's go back and do the slope first.

00:44 So to find out what the slope is, we need d -y -d -t over d -x -d -t.

00:49 So the first thing you're going to do is take the derivative of the y, and the square root, or the derivative of the square root of t is 1 over 2 times the square root of t.

01:02 And then we'll take the derivative of t, which is just one.

01:07 So filling in, we find out that d, y, dx is going to be 1 over 2 square root of t.

01:22 All right, so we'll continue over here to find out what the slope is.

01:27 We're going to evaluate at 1 4th.

01:32 And substituting in 1 4th, we end up with a slope of 1.

01:42 Right to find out what the point is, the value of the point, we're going to go back to the original equations and substitute in one -fourth.

01:57 So when we substitute one -fourth into x of t equals t, we get one -fourth.

02:04 And when we go back in to the original point, y of t equals the square root of t, we'll substitute in one -fourth here...

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