CPT maths

CPT 2559 views 5 replies

Cpt- 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

Replies (5)

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

 

How did we get this2y * dy/dx = 3x^2 * u ?

by differentiation,

and its understood to differentiate, because in the question its given that dy/dx at P = 4

Okay thanks ! I got it!!

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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