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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??

Expert's answer

Question #16187

The gradient of a function $f(x, y) = y - xy + 2px + 3qy$ is $\nabla = (-y + 2p; 1 - x + 3q)$ . Given, that it is equal to $(-2; 3)$ at point $x = 3, y = 2$ , obtain $-2 + 2p = -2; 3q - 2 = 3$ , which gives $p = 0, q = \frac{5}{3}$ .

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Assignment Expert

12.10.12, 17:03

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akansha

12.10.12, 15:59

thanx........:)

Assignment Expert

11.10.12, 15:44

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

akansha

11.10.12, 06:47

sorry will you please ellaborate it, actually i am unable to understand the ans............... i will be thankful to you............ :)