{"id":233085,"date":"2025-06-12T14:26:34","date_gmt":"2025-06-12T14:26:34","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233085"},"modified":"2025-06-12T14:26:36","modified_gmt":"2025-06-12T14:26:36","slug":"given-the-parametric-equations","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/12\/given-the-parametric-equations\/","title":{"rendered":"Given the parametric equations"},"content":{"rendered":"\n

Given the parametric equations:<\/p>\n\n\n\n

\ud835\udc65<\/h1>\n\n\n\n

1
\u2212
\ud835\udc61
2
,<\/p>\n\n\n\n

\ud835\udc66<\/h1>\n\n\n\n

1
+
\ud835\udc61
x=1\u2212t
2
,y=1+t
We are to find:<\/p>\n\n\n\n

\ud835\udc51
\ud835\udc66
\ud835\udc51
\ud835\udc65
dx
dy
\u200b
\u2014 the first derivative of
\ud835\udc66
y with respect to
\ud835\udc65
x,<\/p>\n\n\n\n

\ud835\udc51
2
\ud835\udc66
\ud835\udc51
\ud835\udc65
2
dx
2<\/p>\n\n\n\n

d
2
y
\u200b
\u2014 the second derivative of
\ud835\udc66
y with respect to
\ud835\udc65
x,<\/p>\n\n\n\n

without eliminating the parameter.<\/p>\n\n\n\n

Step 1: First Derivative
\ud835\udc51
\ud835\udc66
\ud835\udc51
\ud835\udc65
dx
dy
\u200b<\/p>\n\n\n\n

Using parametric differentiation, the first derivative is given by:<\/p>\n\n\n\n

\ud835\udc51
\ud835\udc66
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc65<\/h1>\n\n\n\n

\ud835\udc51
\ud835\udc66
\ud835\udc51
\ud835\udc61
\ud835\udc51
\ud835\udc65
\ud835\udc51
\ud835\udc61
dx
dy
\u200b
=
dt
dx
\u200b<\/p>\n\n\n\n

dt
dy
\u200b<\/p>\n\n\n\n

\u200b<\/p>\n\n\n\n

Differentiate
\ud835\udc65
x and
\ud835\udc66
y with respect to
\ud835\udc61
t:<\/p>\n\n\n\n

\ud835\udc51
\ud835\udc65
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc61<\/h1>\n\n\n\n

\ud835\udc51
\ud835\udc51
\ud835\udc61
(
1
\u2212
\ud835\udc61
2<\/p>\n\n\n\n

)<\/h1>\n\n\n\n

\u2212
2
\ud835\udc61
dt
dx
\u200b
=
dt
d
\u200b
(1\u2212t
2
)=\u22122t<\/p>\n\n\n\n

\ud835\udc51
\ud835\udc66
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc61<\/h1>\n\n\n\n

\ud835\udc51
\ud835\udc51
\ud835\udc61
(
1
+
\ud835\udc61<\/p>\n\n\n\n

)<\/h1>\n\n\n\n

1
dt
dy
\u200b
=
dt
d
\u200b
(1+t)=1<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

\ud835\udc51
\ud835\udc66
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc65<\/h1>\n\n\n\n

1
\u2212
2<\/p>\n\n\n\n

\ud835\udc61<\/h1>\n\n\n\n

\u2212
1
2
\ud835\udc61
dx
dy
\u200b
=
\u22122t
1
\u200b
=\u2212
2t
1
\u200b<\/p>\n\n\n\n

Step 2: Second Derivative
\ud835\udc51
2
\ud835\udc66
\ud835\udc51
\ud835\udc65
2
dx
2<\/p>\n\n\n\n

d
2
y
\u200b<\/p>\n\n\n\n

The second derivative in parametric form is given by:<\/p>\n\n\n\n

\ud835\udc51
2
\ud835\udc66
\ud835\udc51
\ud835\udc65<\/p>\n\n\n\n

2<\/h1>\n\n\n\n

\ud835\udc51
\ud835\udc51
\ud835\udc61
(
\ud835\udc51
\ud835\udc66
\ud835\udc51
\ud835\udc65
)
\u00f7
\ud835\udc51
\ud835\udc65
\ud835\udc51
\ud835\udc61
dx
2<\/p>\n\n\n\n

d
2
y
\u200b
=
dt
d
\u200b
(
dx
dy
\u200b
)\u00f7
dt
dx
\u200b<\/p>\n\n\n\n

We already have
\ud835\udc51
\ud835\udc66
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc65<\/h1>\n\n\n\n

\u2212
1
2
\ud835\udc61
dx
dy
\u200b
=\u2212
2t
1
\u200b
, so compute its derivative with respect to
\ud835\udc61
t:<\/p>\n\n\n\n

\ud835\udc51
\ud835\udc51
\ud835\udc61
(
\u2212
1
2
\ud835\udc61<\/p>\n\n\n\n

)<\/h1>\n\n\n\n

1
2
\ud835\udc61
2
dt
d
\u200b
(\u2212
2t
1
\u200b
)=
2t
2<\/p>\n\n\n\n

1
\u200b<\/p>\n\n\n\n

Now divide by
\ud835\udc51
\ud835\udc65
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc61<\/h1>\n\n\n\n

\u2212
2
\ud835\udc61
dt
dx
\u200b
=\u22122t:<\/p>\n\n\n\n

\ud835\udc51
2
\ud835\udc66
\ud835\udc51
\ud835\udc65<\/p>\n\n\n\n

2<\/h1>\n\n\n\n

1
2
\ud835\udc61
2
\u2212
2<\/p>\n\n\n\n

\ud835\udc61<\/h1>\n\n\n\n

\u2212
1
4
\ud835\udc61
3
dx
2<\/p>\n\n\n\n

d
2
y
\u200b
=
\u22122t
2t
2<\/p>\n\n\n\n

1
\u200b<\/p>\n\n\n\n

\u200b
=\u2212
4t
3<\/p>\n\n\n\n

1
\u200b<\/p>\n\n\n\n

Final Answers:
\ud835\udc51
\ud835\udc66
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc65<\/h1>\n\n\n\n

\u2212
1
2
\ud835\udc61
,
\ud835\udc51
2
\ud835\udc66
\ud835\udc51
\ud835\udc65<\/p>\n\n\n\n

2<\/h1>\n\n\n\n

\u2212
1
4
\ud835\udc61
3
dx
dy
\u200b
=\u2212
2t
1
\u200b
,
dx
2<\/p>\n\n\n\n

d
2
y
\u200b
=\u2212
4t
3<\/p>\n\n\n\n

1
\u200b<\/p>\n\n\n\n

Explanation (300 words):
When working with parametric equations, such as<\/p>\n\n\n\n

\ud835\udc65<\/h1>\n\n\n\n

1
\u2212
\ud835\udc61
2
x=1\u2212t
2
and<\/p>\n\n\n\n

\ud835\udc66<\/h1>\n\n\n\n

1
+
\ud835\udc61
y=1+t, we treat both
\ud835\udc65
x and
\ud835\udc66
y as functions of a third variable
\ud835\udc61
t, known as the parameter. Instead of expressing
\ud835\udc66
y directly in terms of
\ud835\udc65
x, we differentiate each with respect to
\ud835\udc61
t, and use these derivatives to find the derivatives of
\ud835\udc66
y with respect to
\ud835\udc65
x.<\/p>\n\n\n\n

To find the first derivative
\ud835\udc51
\ud835\udc66
\ud835\udc51
\ud835\udc65
dx
dy
\u200b
, we apply the chain rule of calculus. Since
\ud835\udc51
\ud835\udc66
\/
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc65<\/h1>\n\n\n\n

(
\ud835\udc51
\ud835\udc66
\/
\ud835\udc51
\ud835\udc61
)
\/
(
\ud835\udc51
\ud835\udc65
\/
\ud835\udc51
\ud835\udc61
)
dy\/dx=(dy\/dt)\/(dx\/dt), we compute the derivatives:<\/p>\n\n\n\n

\ud835\udc51
\ud835\udc66
\/
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc61<\/h1>\n\n\n\n

1
dy\/dt=1, since<\/p>\n\n\n\n

\ud835\udc66<\/h1>\n\n\n\n

1
+
\ud835\udc61
y=1+t,<\/p>\n\n\n\n

\ud835\udc51
\ud835\udc65
\/
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc61<\/h1>\n\n\n\n

\u2212
2
\ud835\udc61
dx\/dt=\u22122t, because<\/p>\n\n\n\n

\ud835\udc65<\/h1>\n\n\n\n

1
\u2212
\ud835\udc61
2
x=1\u2212t
2
.<\/p>\n\n\n\n

Substituting, we get
\ud835\udc51
\ud835\udc66
\ud835\udc51<\/p>\n\n\n\n

\ud835\udc65<\/h1>\n\n\n\n

1
\u2212
2<\/p>\n\n\n\n

\ud835\udc61<\/h1>\n\n\n\n

\u2212
1
2
\ud835\udc61
dx
dy
\u200b
=
\u22122t
1
\u200b
=\u2212
2t
1
\u200b
.<\/p>\n\n\n\n

To find the second derivative
\ud835\udc51
2
\ud835\udc66
\ud835\udc51
\ud835\udc65
2
dx
2<\/p>\n\n\n\n

d
2
y
\u200b
, we use the formula:<\/p>\n\n\n\n

\ud835\udc51
2
\ud835\udc66
\ud835\udc51
\ud835\udc65<\/p>\n\n\n\n

2<\/h1>\n\n\n\n

\ud835\udc51
\ud835\udc51
\ud835\udc61
(
\ud835\udc51
\ud835\udc66
\ud835\udc51
\ud835\udc65
)
\u00f7
\ud835\udc51
\ud835\udc65
\ud835\udc51
\ud835\udc61
dx
2<\/p>\n\n\n\n

d
2
y
\u200b
=
dt
d
\u200b
(
dx
dy
\u200b
)\u00f7
dt
dx
\u200b<\/p>\n\n\n\n

This involves differentiating
\u2212
1
2
\ud835\udc61
\u2212
2t
1
\u200b
with respect to
\ud835\udc61
t, which gives
1
2
\ud835\udc61
2
2t
2<\/p>\n\n\n\n

1
\u200b
, and then dividing this result by
\u2212
2
\ud835\udc61
\u22122t, the derivative of
\ud835\udc65
x. Thus:<\/p>\n\n\n\n

\ud835\udc51
2
\ud835\udc66
\ud835\udc51
\ud835\udc65<\/p>\n\n\n\n

2<\/h1>\n\n\n\n

1
2
\ud835\udc61
2
\u2212
2<\/p>\n\n\n\n

\ud835\udc61<\/h1>\n\n\n\n

\u2212
1
4
\ud835\udc61
3
dx
2<\/p>\n\n\n\n

d
2
y
\u200b
=
\u22122t
2t
2<\/p>\n\n\n\n

1
\u200b<\/p>\n\n\n\n

\u200b
=\u2212
4t
3<\/p>\n\n\n\n

1
\u200b<\/p>\n\n\n\n

This process allows us to compute the derivatives directly from the parametric equations without needing to eliminate the parameter.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Completed Chemistry Worksheet: Exceptions to the Octet Rule<\/strong><\/h3>\n\n\n\n
Molecule or Ion<\/td>Lewis Structure<\/td>Formula Type<\/td>Molecular Shape (Description)<\/td>Shape Name (Molecular Geometry)<\/td>Polarity – if polar show dipoles<\/td><\/tr>
I\u2083\u207b<\/strong><\/td>[ :\u00cf \u2014 \u00cf \u2014 \u00cf: ]\u207b (Central I has 3 lone pairs)<\/td>AX\u2082E\u2083<\/td>The central Iodine atom is bonded to two other Iodine atoms in a straight line, with three lone pairs arranged in a plane around the central atom’s equator.<\/td>Linear<\/strong><\/td>Nonpolar<\/strong> (The symmetrical arrangement of atoms causes the dipoles to cancel.)<\/td><\/tr>
PCl\u2086\u207b<\/strong><\/td>[P bonded to 6 Cl]\u207b (P has no lone pairs)<\/td>AX\u2086<\/td>A central Phosphorus atom is bonded to six Chlorine atoms, with four in a square plane and one above and one below the plane.<\/td>Octahedral<\/strong><\/td>Nonpolar<\/strong> (The symmetrical octahedral geometry causes all bond dipoles to cancel out.)<\/td><\/tr>
SF\u2084<\/strong><\/td>S bonded to 4 F (S has one lone pair)<\/td>AX\u2084E\u2081<\/td>A central Sulfur atom is bonded to four Fluorine atoms. The lone pair on the sulfur atom pushes the bonds away, creating a shape that resembles a seesaw.<\/td>Seesaw<\/strong><\/td>Polar<\/strong> (The shape is asymmetrical due to the lone pair. There is a net dipole moment pointing towards the fluorine atoms.)<\/td><\/tr>
BF\u2083<\/strong><\/td>B bonded to 3 F (B has an incomplete octet)<\/td>AX\u2083<\/td>A central Boron atom is bonded to three Fluorine atoms, all lying in the same plane and spaced 120\u00b0 apart.<\/td>Trigonal Planar<\/strong><\/td>Nonpolar<\/strong> (The symmetrical arrangement of the B-F bonds in a plane causes the dipoles to cancel.)<\/td><\/tr>
ClF\u2083<\/strong><\/td>Cl bonded to 3 F (Cl has two lone pairs)<\/td>AX\u2083E\u2082<\/td>A central Chlorine atom is bonded to three Fluorine atoms. Two lone pairs on the chlorine push the bonds into a T-shape.<\/td>T-shaped<\/strong><\/td>Polar<\/strong> (The shape is asymmetrical. The bond dipoles do not cancel, resulting in a net dipole moment along the stem of the ‘T’.)<\/td><\/tr>
ClF\u2084\u207b<\/strong><\/td>[Cl bonded to 4 F]\u207b (Cl has two lone pairs)<\/td>AX\u2084E\u2082<\/td>A central Chlorine atom is bonded to four Fluorine atoms that lie in a single plane. Two lone pairs are located above and below this plane.<\/td>Square Planar<\/strong><\/td>Nonpolar<\/strong> (Although the bonds are polar, the symmetrical arrangement of the four Fluorine atoms around the central Chlorine causes their dipoles to cancel out.)<\/td><\/tr>
XeF\u2082<\/strong><\/td>Xe bonded to 2 F (Xe has three lone pairs)<\/td>AX\u2082E\u2083<\/td>A central Xenon atom is bonded to two Fluorine atoms. Three lone pairs on the Xenon atom are arranged in a plane, forcing the Fluorine atoms to opposite sides.<\/td>Linear<\/strong><\/td>Nonpolar<\/strong> (The two Xe-F bonds are on opposite sides of the central atom, so their dipoles are equal and opposite, and they cancel out.)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

thumb_upthumb_down<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

Given the parametric equations: \ud835\udc65 1\u2212\ud835\udc612, \ud835\udc66 1+\ud835\udc61x=1\u2212t2,y=1+tWe are to find: \ud835\udc51\ud835\udc66\ud835\udc51\ud835\udc65dxdy\u200b\u2014 the first derivative of\ud835\udc66y with respect to\ud835\udc65x, \ud835\udc512\ud835\udc66\ud835\udc51\ud835\udc652dx2 d2y\u200b\u2014 the second derivative of\ud835\udc66y with respect to\ud835\udc65x, without eliminating the parameter. Step 1: First Derivative\ud835\udc51\ud835\udc66\ud835\udc51\ud835\udc65dxdy\u200b Using parametric differentiation, the first derivative is given by: \ud835\udc51\ud835\udc66\ud835\udc51 \ud835\udc65 \ud835\udc51\ud835\udc66\ud835\udc51\ud835\udc61\ud835\udc51\ud835\udc65\ud835\udc51\ud835\udc61dxdy\u200b=dtdx\u200b dtdy\u200b \u200b Differentiate\ud835\udc65x and\ud835\udc66y with respect to\ud835\udc61t: […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-233085","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233085","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=233085"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233085\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233085"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233085"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233085"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}