   Chapter 10.2, Problem 24E

Chapter
Section
Textbook Problem

# Graph the curve in a viewing rectangle that displays all the important aspects of the curve.24. x = r4 + 4t3 − 8t2, y = 2t2 − t

To determine

To plot: The curve for the given equation parametric equations x=t4+4t38t2 and y=2t2t .

Explanation

Given:

The parametric equation for the variable x is as follows.

x=t4+4t38t2 (1)

The parametric equation for the variable y is as follows.

y=2t2t (2)

Calculation:

Differentiate the parametric equation (1) for the variable x .

dxdt=t4+4t38t2=4t+12t216tdxdt=4t+12t216t (3)

Differentiate the parametric equation (2) for the variable y .

dydt=2t2t

dydt=4t1 (4)

Substitute 1 for t in equation (1).

x=t4+4t38t2=(1)4+4(1)38(1)2x=1+48x=3

Substitute 1 for t in equation (2).

y=2t2t=2(1)2(1)y=21

y=1

The values of x and y for each step value of t is tabulated in the below table.

 t 0 1 2 3 x 0 −3 0 45 y 0 1 7 17

Graph:

The loop graph lies in the interval (3,0) .

The horizontal and vertical tangents are obtained as shown below in figure 2.

Horizontal tangents can be drawn to the curve at the coordinate point (0.4,0.1) .

Vertical tangents can be drawn to the curve at the coordinate points (3,1) and (0,0) .

Write the chain rule for dydx .

dydx=dydtdxdtdydx=4t14t+12t216t

Use the condition below for horizontal tangents.

dydt=04t1=0t=14

Substitute 14 for t in equation (1).

x=t4+4t38t2=(14)4+4(14)38(14)2x=0.4335

Substitute 14 for t in equation (2)

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