Find the equation of the line tangent to the curve at the point defined by the given value of t.

step 1 In this question, we will be using the chain rule. The question gives us a parametrically define

Find the equation of the line tangent to the curve at the...

Key Concepts

Tangent Line Equation

Once the slope of the tangent line at a given point is determined, the equation of the tangent line can be written using point-slope form. This involves substituting the coordinates of the point and the calculated slope into the formula y - y1 = m(x - x1). This linear equation approximates the curve near the point of tangency and provides insight into the local behavior of the curve.

Slope of the Tangent Line

The slope of a tangent line to a curve represents the rate of change of the curve at a specific point and is calculated by taking the derivative, often using the parametric derivatives as described above. The computed slope is a critical component in determining the behavior of the curve at that point, as it directly influences the line's angle and direction.

Parametric Equations

Parametric equations define a relationship between two variables, typically x and y, using a third parameter, often denoted as t. This representation is especially useful for describing curves that are not functions in the conventional sense, and it allows for the independent control of each coordinate, which can simplify the analysis of curves, particularly when dealing with periodic or circular motion.

Differentiation of Parametric Equations

Differentiating parametric equations involves taking the derivative of x and y with respect to the parameter and then forming the ratio (dy/dt)/(dx/dt) to obtain dy/dx. This process finds the slope of the tangent to the curve at a given parameter value and is a standard method in calculus for analyzing the instantaneous rate of change along a curve defined parametrically.

Transcript

00:01 In this question, we will be using the chain rule.

00:06 The question gives us a parametrically defined curve and asks us to find a tangent line to the curve at the point with t is equal to pi over 4.

00:18 That is, the parameter is pi over 4, and that gives us the x and y coordinates.

00:31 So, first of all, what is the formula for the tangent line? well, given that the point a, b is the point of tangency, where a is the x coordinate, and b is the y coordinate, the quantities y minus b and x minus a are related by the slope.

01:01 That is y minus b divided by x minus a is equal to the derivative at the point we are considering.

01:10 At the value of t we are considering.

01:16 So we can easily find the x coordinate just by substituting the t value into x and the y coordinate follows the same path.

01:31 We only need to find the b y d x term.

01:36 And so let's go about doing that first.

01:41 First, let's find d y d x, and then we'll evaluate it at t equals pi over four.

01:50 So by the chain rule, we know that d, y, dx is d ydt over dx d t, as long as dx dt is non -z.

02:07 And by the way, that comes from this form of the chain rule that we are used to most.

02:16 Okay.

02:17 Now that we have that, we can find d, y, dx by first differentiating y with respect to t, and then x, and then dividing the results...

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