CA Foundation Exam > CA Foundation Questions > The curve y2 = ux3 v passes through the poi...
Start Learning for Free
The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are?
Most Upvoted Answer
The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 ...
Given Information:
- The curve is represented by the equation y^2 = ux^3 + v.
- The curve passes through the point P(2, 3).
- The derivative dy/dx = 4 at point P.
Step 1: Finding the values of u and v
To find the values of u and v, we need to substitute the coordinates of point P into the equation of the curve and solve for u and v.
Substituting point P(2, 3) into the equation:
(3)^2 = u(2)^3 + v
9 = 8u + v
Step 2: Finding the derivative of the equation
To find the derivative dy/dx, we need to differentiate the equation of the curve with respect to x.
Differentiating the equation:
d/dx(y^2) = d/dx(ux^3 + v)
2y(dy/dx) = 3ux^2
Step 3: Substituting the values of x and y at point P
We are given that dy/dx = 4 at point P(2, 3). Substituting these values into the derivative equation, we can solve for u.
Substituting point P(2, 3) into the derivative equation:
2(3)(4) = 3u(2)^2
24 = 12u
u = 2
Step 4: Solving for v
Now that we have the value of u, we can substitute it back into the equation obtained in Step 1 to solve for v.
Substituting u = 2 into the equation:
9 = 8(2) + v
9 = 16 + v
v = -7
Step 5: Final answer
The values of u and v are found to be u = 2 and v = -7, respectively.
Therefore, the curve is represented by the equation y^2 = 2x^3 - 7.
- The curve is represented by the equation y^2 = ux^3 + v.
- The curve passes through the point P(2, 3).
- The derivative dy/dx = 4 at point P.
Step 1: Finding the values of u and v
To find the values of u and v, we need to substitute the coordinates of point P into the equation of the curve and solve for u and v.
Substituting point P(2, 3) into the equation:
(3)^2 = u(2)^3 + v
9 = 8u + v
Step 2: Finding the derivative of the equation
To find the derivative dy/dx, we need to differentiate the equation of the curve with respect to x.
Differentiating the equation:
d/dx(y^2) = d/dx(ux^3 + v)
2y(dy/dx) = 3ux^2
Step 3: Substituting the values of x and y at point P
We are given that dy/dx = 4 at point P(2, 3). Substituting these values into the derivative equation, we can solve for u.
Substituting point P(2, 3) into the derivative equation:
2(3)(4) = 3u(2)^2
24 = 12u
u = 2
Step 4: Solving for v
Now that we have the value of u, we can substitute it back into the equation obtained in Step 1 to solve for v.
Substituting u = 2 into the equation:
9 = 8(2) + v
9 = 16 + v
v = -7
Step 5: Final answer
The values of u and v are found to be u = 2 and v = -7, respectively.
Therefore, the curve is represented by the equation y^2 = 2x^3 - 7.
Attention CA Foundation Students!
To make sure you are not studying endlessly, EduRev has designed CA Foundation study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in CA Foundation.
|
Explore Courses for CA Foundation exam
|
|
Top Courses for CA FoundationView all
The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are?
Question Description
The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are?.
The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are? for CA Foundation 2024 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are? covers all topics & solutions for CA Foundation 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are?.
Solutions for The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are? in English & in Hindi are available as part of our courses for CA Foundation.
Download more important topics, notes, lectures and mock test series for CA Foundation Exam by signing up for free.
Here you can find the meaning of The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are? defined & explained in the simplest way possible. Besides giving the explanation of
The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are?, a detailed solution for The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are? has been provided alongside types of The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are? theory, EduRev gives you an
ample number of questions to practice The curve y2 = ux3 v passes through the point P(2, 3) and dy dx = 4 at P. The values of u and v are? tests, examples and also practice CA Foundation tests.
|
|
Explore Courses for CA Foundation exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup