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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?
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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.
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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?
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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?.
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