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

Key Concepts

Parametric Equations

Parametric equations describe a curve by expressing the coordinates as functions of an independent parameter. This approach allows for the representation of complex curves that might be difficult or impossible to express as a single function in Cartesian coordinates.

Derivative of Parametric Equations

For curves given in parametric form, the derivative dy/dx is found by taking the derivative of y with respect to the parameter and dividing it by the derivative of x with respect to the parameter. This ratio provides the slope of the tangent line at a specific point on the curve.

Tangent Line to a Curve

The tangent line to a curve at a given point is the line that just touches the curve at that point. Its slope is given by the derivative of the curve evaluated at that point, and its equation can be formed using the point-slope formula.

Chain Rule in Parametric Differentiation

The chain rule is essential in parametric differentiation. It allows the computation of dy/dx by linking the derivatives with respect to the parameter, and it is also used to determine higher-order derivatives such as the second derivative with respect to x.

Second Derivative in Parametric Form

To find the second derivative d²y/dx² for a curve defined parametrically, one differentiates the first derivative dy/dx with respect to the parameter and then divides by the derivative of x with respect to the parameter. This process gives insight into the curvature or concavity of the curve at a point.

Transcript

00:01 So in this question, we want to find an equation for the line tangent to the curve at the point defined by the given t value.

00:07 Also, we're going to find our second derivative at that point.

00:12 Here i have x equals the secant squared of t minus 1, and y equals the tangent of t.

00:18 I'm at t equals negative pi over 4.

00:22 Now if i want an equation of my tangent line, i need two pieces of information.

00:27 I need a point and a slope.

00:29 So let's get my point.

00:33 Let's get my x.

00:37 So my x is the secant squared of negative pi over 4 minus 1.

00:47 Well, what do i know that will help me figure that out? well, first of all, i know that the cosine of negative pi over 4 is equal to square root of 2 over 2.

01:01 That means that the secant of negative pi over 4 is equal to 2 over root 2 or square root 2.

01:15 That implies that the secant squared of negative pi over 4, when i square that root 2, i just get 2 minus 1 equals 1.

01:28 While my y, it is the tangent of negative pi over 4.

01:35 Tangent of negative pi over 4 is negative 1.

01:39 So my point is 1 comma negative 1.

01:45 Now let's get my slope.

01:47 So to get my slope, i need dy dx.

01:52 To get dy dx, i take the derivative of y with respect to t, and i divide that by the derivative of x with respect to t.

02:07 My derivative of y with respect to t, the derivative of tangent t, is secant squared of t.

02:15 The derivative of x with respect to t, the derivative of secant squared of t using the chain rule is 2 times the secant of t times the secant of t tangent of t.

02:34 Now this is going to simplify my secant squared of t cancels with my factors of secant t in the denominator.

02:45 So that my dy dx looks like 1 over 2 tangent of t, or equivalently, 1 half times the cotangent of t.

03:03 Now to get my slope, i have to plug in my t value, negative pi over 4.

03:13 So i'm getting 1 half the cotangent of negative pi over 4.

03:22 Remember the tangent of negative pi over 4 was negative 1.

03:26 So the cotangent of negative pi over 4 is also negative 1...

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