Please answer parts iv, v and vi, considering the attached answers for i, ii and iii. PART iThe y axis intersection is (0,0) and (0,2)Therefore x axis intersection is at (0,0).PART iiHence the critical point is at (2,4).PART iiiThe stationary point is a local maximum Consider the parametric curve given byx(t)t² + ty(t) = t³ + 3t²=}te [-π,5/4].(i) Find the axis intercepts.(ii) Find the critical points.(iii) Characterise the turning points.(iv) Find the extreme points.(v) Sketch the curve ensuring the direction of motion is indicated and all major points (from (i)–(iv))are labelled.(vi) Find the equation of the line tangent to the curve when t = 1.

The values of .The slope the tangent at is the derivative of the given curve at .The derivative of…

Answered: (iv) Find the extreme points. (v)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Please answer parts iv, v and vi, considering the attached answers for i, ii and iii.

PART i
The y axis intersection is (0,0) and (0,2)
Therefore x axis intersection is at (0,0).
PART ii
Hence the critical point is at (2,4).
PART iii
The stationary point is a local maximum

Consider the parametric curve given by
x(t)
t² + t
y(t) = t³ + 3t²
=
}
te [-π,5/4].
(i) Find the axis intercepts.
(ii) Find the critical points.
(iii) Characterise the turning points.
(iv) Find the extreme points.
(v) Sketch the curve ensuring the direction of motion is indicated and all major points (from (i)–(iv))
are labelled.
(vi) Find the equation of the line tangent to the curve when t = 1.

Expert Solution

Step 1: The derivative of a parametric function

The values of $x open parentheses t close parentheses space a n d space y open parentheses t close parentheses space a t space t equals 1 space space$ .

$x open parentheses 1 close parentheses equals 1 squared plus 1 equals 2 y open parentheses 1 close parentheses equals 1 cubed plus 3 open parentheses 1 close parentheses squared equals 1 plus 3 equals 4$

The slope the tangent at $t$ is the derivative of the given curve at $t$ .

The derivative of a parametric function is given by

$fraction numerator d y over denominator d x end fraction equals fraction numerator fraction numerator d y over denominator d t end fraction over denominator fraction numerator d x over denominator d t end fraction end fraction equals fraction numerator 3 t squared plus 6 t over denominator 2 t plus 1 end fraction equals 9 over 3 equals 3 space a t space space t equals 1$