Tangent lines Find an equation of the line tangent to the curve at the point corresponding to th

Tangent lines Find an equation of the line tangent to the...

Key Concepts

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter. This method allows complex curves to be described in terms of simpler, often more intuitive functional relationships, and is commonly used in physics and engineering to model motion and change.

Derivative of a Parametric Curve

When a curve is defined parametrically, the slope of the tangent line at any point is found by differentiating the coordinate functions with respect to the parameter. The derivative dy/dx is computed as (dy/dt) divided by (dx/dt), which provides a measure of the rate at which the y-coordinate changes relative to the x-coordinate at a given parameter value.

Tangent Line to a Curve

The tangent line at a point on a curve is the line that just touches the curve at that point, reflecting its instantaneous direction. Finding this line usually involves determining the point on the curve and the slope of the curve at that point, then applying the point-slope form of a line.

Evaluating at a Specific Parameter Value

To obtain the exact point and slope of a tangent line, the parametric equations and their derivatives are evaluated at a specific value of the parameter. This substitution provides the coordinates of the point of tangency and the corresponding slope, which are then used to formulate the equation of the tangent line.

Transcript

00:01 The parametric equation is x is equal to cosine t plus t sine of t and y is equal to sine of t minus t cosine of t.

00:18 And we want to find the equation of tangent where t is equal to pi over four.

00:33 So let's first of all start by finding the derivative.

00:38 So we have dx over dt.

00:42 That would be equal to negative sign of t plus t cosine of t.

00:51 That's a product rule plus sign of t.

00:56 And that simplifies to t cosine of t.

01:03 Similarly, we can find d .y by dt.

01:07 So d y by dt would be the derivative of.

01:10 Of t which is a cosine of t minus the derivative of t cosine t so use the product rule that would be first function times the derivative of the second function plus second function times the derivative of the first function which will just be one so so simplifying this gives cosine of t plus t sine of t minus cosine of t.

01:49 So d y by d t is equal to t times sine of t.

01:55 Now that we have d x by d t and d y by d t we can find d y by d x, which is equal to d -y by d -t divided by d -x by d -t and that would be equal to t of sine t divided by t times cosine of t and that simplifies to tangent of t.

02:19 So we can now find d -y by d -x for t is equal to pi over four and that would be equal to tangent of pi over four which is equal to 1.

02:33 So we now have our slope.

02:36 So we know that the slope for the tangent would be equal to 1.

02:41 Now we need a point.

02:44 So for finding the point, we just need to find the value of x where t is equal to pi over 4...

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