alfiya (student) (133 Points)
13 May 2015Cpt- i need help in this maths question. being a non-maths student i am finding it difficult to solve questions that require more of formulas and steps. i would be very thankful if someone could explain to me how to solve this question
1.The curve y^2=ux^3+v passes through the point P(2,3) and dy/dx=4 at P. The values of u and v are: A. 2,7 B. 2,-7 C. -2,-7 D. 0,-1
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 13 May 2015
y^2 = u*x^3 + v
differentiating w.r.t x
2y * dy/dx = 3x^2 * u
therefore, dy/ dx = 3x^2 * u / 2y
substituing (x,y) = (2,3)
dy/dx = 3 * 2 * 2 * u / 2 * 3
4 = 2u
therefore u = 2
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 13 May 2015
by differentiation,
and its understood to differentiate, because in the question its given that dy/dx at P = 4
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 13 May 2015
now put u=2 in
y^2 = u.x^3 + v
and substitute (x,y) = (2,3) only because it is mentioned in the question that the curve
y^2 = u.x^3 + v passes through the point (2,3)
3^2 = 2 * 2^3 + v
9 = 16 + v
therefore v = 9 - 16
= - 7
therefore ----- u =2 & v = -7
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