Question

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. $$ x=\sqrt{t}, \quad y=t^{2}-2 t ; \quad t=4 $$

Name: Problem 8, Chapter 10: Calculus: Early Transcendentals, Metric Edition
Uploaded: 2022-09-19T16:25:40-08:00
Channel: Carson Merrill
Description: Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x=√(t), y=t^2-2 t ; t=4

   Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter.
$$
x=\sqrt{t}, \quad y=t^{2}-2 t ; \quad t=4
$$

Calculus: Early Transcendentals, Metric Edition

James Stewart,… 9th Edition

Chapter 10, Problem 8

Instant Answer

Solved by Expert Carson Merrill

09/19/2022

Step 1

The derivative of x with respect to t is given by: \[dx/dt = 1/(2\sqrt{t})\] And the derivative of y with respect to t is given by: \[dy/dt = 2t - 2\] Show more…

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Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. $$ x=\sqrt{t}, \quad y=t^{2}-2 t ; \quad t=4 $$

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Key Concepts

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of an independent parameter. They are useful for representing curves that may not be functions in y = f(x) form, and they allow complex geometric shapes to be described more flexibly.

Tangent Line

The tangent line to a curve at a particular point is the line that just touches the curve at that point and has the same instantaneous direction. It represents the best linear approximation of the curve near that point.

Derivative of a Parametric Curve

For curves defined parametrically, the slope of the tangent line is obtained by differentiating the coordinate functions with respect to the parameter and then computing dy/dx as (dy/dt)/(dx/dt). This process is essential for determining the direction of the tangent and performing further analysis on the curve.

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