Home
NCERT Soltions
Class 12
Mathematics
Application of Derivatives
A particle moves along the curve 6y = x3 + 2. Find the points on the c...

A particle moves along the curve 6y = x3 | Class 12 Mathematics Chapter Application of Derivatives, Application of Derivatives NCERT Solutions

Question:

A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Answer:

The equation of the curve is given as:

6y = x³ + 2

The rate of change of the position of the particle with respect to time (t) is given by,

\begin{align}6\frac{dy}{dt} = 3x^2\frac{dx}{dt}+0\end{align}

\begin{align}\Rightarrow 2\frac{dy}{dt} = x^2\frac{dx}{dt}\end{align}

When the y-coordinate of the particle changes 8 times as fast as the

\begin{align}x-coordinate\; i.e.,\left(\frac{dy}{dt} = 8\frac{dx}{dt}\right), we \;have:\end{align}

\begin{align}2\left(8.\frac{dx}{dt}\right) = x^2.\frac{dx}{dt}\end{align}

\begin{align}\Rightarrow 16.\frac{dx}{dt} = x^2.\frac{dx}{dt}\end{align}

\begin{align}\Rightarrow (x^2 - 16).\frac{dx}{dt} =0 \end{align}

\begin{align}\Rightarrow x^2=16 \end{align}

\begin{align}\Rightarrow x=\pm 4 \end{align}

\begin{align}When\; x = 4, y = \frac{4^3 + 2}{6}=\frac{66}{6}=11\end{align}

\begin{align}When\; x = -4, y = \frac{(-4)^3 + 2}{6}=\frac{-62}{6}=\frac{-31}{3}\end{align}

Hence, the points required on the curve are

\begin{align} (4,11)\; and \;(-4,\frac{-31}{3}).\end{align}

1
0
0
Report?

Report a problem on Specifications:

Email id:

Comments

Next Question >

Popular questions of class 12 Mathematics

Q: Given an example of a relation. Which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.

Q: Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A = {1, 2, 3,13, 14} defined as
R = {(x, y): 3x − y = 0}
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y): y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x − y is as integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y): x and y work at the same place}
(b) R = {(x, y): x and y live in the same locality}
(c) R = {(x, y): x is exactly 7 cm taller than y}
(d) R = {(x, y): x is wife of y}
(e) R = {(x, y): x is father of y}

Q: Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Q: Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

Q: Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive.

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i) f : R → R defined by f(x) = 3 – 4x

(ii) f : R → R defined by f(x) = 1 + x²

Q: Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.

Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Q: Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.

Recently Viewed Questions of Class 12 Mathematics

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i) f : R → R defined by f(x) = 3 – 4x

(ii) f : R → R defined by f(x) = 1 + x²

Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Q: Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

Q: \begin{align} \int\left(1-x\right).\sqrt {x}.dx\end{align}

Q: Integrals (ax + b)²

Consider f : R₊ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f^–1 of f given by , where R₊ is the set of all non-negative real numbers.

Determine order and degree(if defined) of differential equation \begin{align}\left(\frac{ds}{dt}\right)^4\;+\;3s\frac{d^2s}{dt^2}\;=\;0\end{align}

A balloon, which always remains spherical, has a variable diameter

\begin{align} \frac{3}{2}(2x+1)\end{align}

Find the rate of change of its volume with respect to x.

Q: Integrals sin 2x

Determine order and degree(if defined) of differential equation

\begin{align}\left(\frac{d^2y}{dx^2}\right)^2\;+\;cos\left(\frac{dy}{dx}\right)\;=\;0\end{align}

Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

Read the question carefully and focus on the core concept being asked.
Reference examples and data from the chapter when answering questions about Application of Derivatives.
Review previous year question papers to get an idea of how such questions may be framed in exams.
Practice answering questions within the time limit to improve your speed and accuracy.
Discuss your answers with your teachers or peers to get feedback and improve your understanding.

Comments

Comment(s) on this Question

Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 11: A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y-coordinate....

A particle moves along the curve 6y = x3 | Class 12 Mathematics Chapter Application of Derivatives, Application of Derivatives NCERT Solutions

< Previous Question

Next Question >

Popular questions of class 12 Mathematics

Recently Viewed Questions of Class 12 Mathematics

Study Tips for Answering NCERT Questions:

Comments

Comment(s) on this Question

Chapters for Class 12 Mathematics

Subjects for Class 12

Students Also Read

Important links

Quick Links

Resources

Support