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Ch. 11 - Parametric Equations and Polar Coordinates
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 11, Problem 11.2.6

Tangent Lines to Parametrized Curves


In Exercises 1−14, 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 d²y/dx² at this point.


x = sec² t − 1, y = tan t, t = −π/4

Verified step by step guidance
1
First, find the derivatives of the parametric equations with respect to the parameter \(t\). Compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) from the given functions \(x = \sec^{2} t - 1\) and \(y = \tan t\).
Next, find the slope of the tangent line \(\frac{dy}{dx}\) at the given value of \(t = -\frac{\pi}{4}\) by using the chain rule for parametric curves: \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
Evaluate \(x\) and \(y\) at \(t = -\frac{\pi}{4}\) to find the specific point on the curve where the tangent line touches.
Use the point-slope form of a line, \(y - y_0 = m(x - x_0)\), where \(m\) is the slope found in step 2 and \((x_0, y_0)\) is the point found in step 3, to write the equation of the tangent line.
To find \(\frac{d^{2}y}{dx^{2}}\) at \(t = -\frac{\pi}{4}\), first compute \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\), then divide by \(\frac{dx}{dt}\) using the formula \(\frac{d^{2}y}{dx^{2}} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\), and finally evaluate at \(t = -\frac{\pi}{4}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Parametric Equations and Curves

Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted t. Understanding how x and y depend on t allows us to analyze the curve's shape and behavior by varying t, rather than expressing y directly as a function of x.
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Finding the Tangent Line to a Parametric Curve

The tangent line at a point on a parametric curve is found using derivatives dx/dt and dy/dt. The slope of the tangent line is dy/dx = (dy/dt) / (dx/dt), evaluated at the given t. Using this slope and the point coordinates, the tangent line equation can be written in point-slope form.
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Differentiation of Parametric Curves

Second Derivative d²y/dx² for Parametric Curves

The second derivative d²y/dx² measures the curvature of the parametric curve and is found by differentiating dy/dx with respect to x. Using the chain rule, d²y/dx² = (d/dt(dy/dx)) / (dx/dt). Evaluating this at the given t provides the curve's concavity at that point.
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Higher Order Derivatives of Parametric Curves