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# Answer to Question #16187 in Calculus for akansha

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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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............ :)

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