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Calculus / Calculus: Early Transcendentals 1 / Chapter 10.1 / Problem 45E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

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Derivatives Consider the following parametric curves.a.

Chapter 8, Problem 45E

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QUESTION:

Consider the following parametric curves.

a. Determine dy/dx in terms of t and evaluate it at the given value of t.

b. Make a sketch of the curve showing the tangent line at the point corresponding to the given value of t.

x = 2 + 4t, y = 4 - 8t; t = 2

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QUESTION:

Consider the following parametric curves.

a. Determine dy/dx in terms of t and evaluate it at the given value of t.

b. Make a sketch of the curve showing the tangent line at the point corresponding to the given value of t.

x = 2 + 4t, y = 4 - 8t; t = 2

ANSWER:

Solution 45E

Step 1:
        To obtain the derivative we need $\frac{dy}{dx}$ which we can also write as
         $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
        Evaluating $\frac{dy}{dt}$ and $\frac{dx}{dt}$ separately, we get

$\frac{dy}{dt}=\frac{d(4-8t)}{dt}$
$=-8$

$\frac{dx}{dt}=\frac{d(2+4t)}{dt}$
$=4$
Putting the values $\frac{dy}{dt}$ and $\frac{dx}{dt}$ in $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ to evaluate $\frac{dy}{dx}$ , we get

$\frac{dy}{dx}=\frac{-8}{4}\Rightarrow{}-2$ .
Hence the value of $\frac{dy}{dx}$ is $-2$ which is independent of t.

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Derivatives Consider the following parametric curves.a.

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Solution Found!

Derivatives Consider the following parametric curves.a.

Questions & Answers

Study Tools You Might Need

Chapter: 9 Problem: 4

Chapter: 11 Problem: 11.12

Chapter: 44 Problem: 1

Chapter: 43 Problem: 3

Chapter: 14 Problem: 64

Chapter: 13 Problem: 1

Chapter: 2 Problem: 28

Chapter: 9 Problem: 30

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