Quantitative aptitude
This query is : Resolved
28 January 2014
How to solve this problem
The gradient of the curve y minus xy plus 2px plus 3qy is equal to 0 at the point (3,2)is minus 2/3, the values of p and q are...
The gradient of the curve y minus xy plus 2px plus 3qy is equal to 0 at the point (3,2)is minus 2/3, the values of p and q are...
29 January 2014
ans. p=1/2 q=1/6
Problem related to caluculas
only point to remeber is slope means dy/dx
and formula of dy/dx(xy)
solution:
Differentiating the given equation, we get
dy/dx - xdy/dx - y + 2p + 3qdy/dx = 0.
If we put the value of slope = dy/dx = -2/3,
the value of x = 3 and the value of y = 2, then we get:
(-2/3) - 3(-2/3) - (2) + 2p + 3q(-2/3) = 0.
So,3p-3q= 1
3p=1+3q. ....(1)
the point (3, 2) passes through the curve given.
Therefore, 2 - (3)(2) + 2(p)(3) + 3(q)(2) = 0. So, 3p + 3q = 2.......(2)
substitute (1) in (2)
we get q = 1/6.
keep q= 1/6 in (1) or (2)
.
can get p = 1/2
Problem related to caluculas
only point to remeber is slope means dy/dx
and formula of dy/dx(xy)
solution:
Differentiating the given equation, we get
dy/dx - xdy/dx - y + 2p + 3qdy/dx = 0.
If we put the value of slope = dy/dx = -2/3,
the value of x = 3 and the value of y = 2, then we get:
(-2/3) - 3(-2/3) - (2) + 2p + 3q(-2/3) = 0.
So,3p-3q= 1
3p=1+3q. ....(1)
the point (3, 2) passes through the curve given.
Therefore, 2 - (3)(2) + 2(p)(3) + 3(q)(2) = 0. So, 3p + 3q = 2.......(2)
substitute (1) in (2)
we get q = 1/6.
keep q= 1/6 in (1) or (2)
.
can get p = 1/2
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