Question #16187
the gradient of the curve y-xy+2px+3qy=0 at the point (3,2) is -2/3 the value of p & q are??
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First of all, we need to find expression for gradient. Gradient is a vector which components are partial derivatives of the curve. Having curve f(x,y)=y-xy+2px+3qy=0& we can find its gradient as follows: gradx=partial derivative of f by x=-y+2p grady=partial derivative of f by y=1-x+3q So the gradient is grad(x,y)=(-y+2p, 1-x+3q) Then, we're given that at the point (x=3,y=2) gradient equals (-2,3) At first, let's find gradient at the point x=3,y=2. Which is grad(3,2)=(-2+2p, 1-3+3q) Now we equate obtained components to given (-2,3): gradx=-2+2p=-2 grady=1-3+3q=3 From these equations we find p and q Hopefully this will help
sorry will you please ellaborate it, actually i am unable to understand the ans............... i will be thankful to you............ :)