find the slope of the tangent to the curve xy^2 +x^2y=2 at the point (1,1) ?

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jkbuck

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find the slope of the tangent to the curve xy^2 +x^2y=2 at the point (1,1)

I need an answer to that. Thank you

Posted 227 day ago

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Answers (4)

jayteacher

The slope problem is calculated normally with a graphing calculator or with implicit differentiation.
dy/dx[xy^2] + dy/dx[x^2y] = dy/dx[2]
x(2y)(dy/dx) + (y^2)(dx/dx) + (2xy)(dx/dx) + (x^2)(dy/dx) = 0.
x(2y)(dy/dx) + (y^2) + (2xy) + (x^2)(dy/dx) = 0.
2xy(dy/dx) + x^2(dy/dx) = – (y^2) – 2xy.
(2xy +x^2)(dy/dx) = – (2xy + y^2).
dy/dx = – (2xy + y^2) / (2xy + x^2.)

For dy/dx at (1, 1), substitute x = 1 and y = 1.
dy/dx = – [2(1)(1) + (1)^2] / [2(1)(1) + (1)^2]
dy/dx = – [3] / [3] = – 1

Posted 224 day ago

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MFB13

This equation is called an implicit second degree equation. This is because y is not necessarily a simple function of x such that y = f(x). Thus an equation like the example you gave may imply that one of the variables can be represented as a function, or combination of functions, of the other which is refered to as implicit functions. By using differential calculus to differentiate this mixed function implicitly, you do not have to separate the variables before differentiating. But at the end, you will be able to solve for the derivative or instantaineous slope of the entire expresion in terms of x and y. Let us first write this implicit funtion as given:

xy^2 + x^2y = 2

Next, we differentiate this function implicitly by using both the product rule and the chain rule for differentiation: That is as follows:

1(Y^2) + X(2Y)(dy/dx) + 2X(Y)+X^2(dy/dx) = 0

We now place like terms that include dy/dx on one side of the equation and transpose the other terms to the other side of the equation giving the following expression:

2YX(dy/dx) + X^2(dy/dx) = -Y^2 - 2XY

Next, we factor out the dy/dx on the left side and the minus signs on the right side as follows:

dy/dx (2YX + X^2) = - (Y^2 + 2XY)

We now solve for dy/dx by dividing both sides by (2YX + X^2) :
dy/dx = - (Y^2 + 2XY)/(2YX + X^2)

The last step we have to do is substitute the X value and Y value from the point, (1,1) into the above equation where X = 1 and Y = 1 as follows:

dy/dx = -[1^2 + 2(1)(1)] / [2(1)(1) + 1^2] = -(1+2) / (2+1) = -3/3 = - 1

Posted 227 day ago

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atta

The above equation entails a special type of differentiation known as implicit-differentiation., which entails that both sides of the equation be differentiated when there x and y terms involved. The line tangent to the curve at the specified point of (1, 1) entails elementary algebraic methods for finding the slope of the corresponding line. Therefore, the slope of the line can be found by using Calculus, which is as follows:

dy/dx[xy^2] + dy/dx[x^2y] = dy/dx[2] => x2ydy/dx + y^2(1) + 2xy(1) + x^22dy/dx = 0 => x2ydy/dx + x^2dy/dx = -y^2 - 2xy => dy/dx[] = -y^2 - 2xy => dy/dx[2xy +x^2 ] = -2xy - y^2 => dy/dx = -2xy - y^2 / 2xy + x^2. Have found dy/dx, all you need to do is find the point-slope form and substitute.

J.C

Posted 227 day ago

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EmmaLam

Use implicit differentiation:

xy^2+x^2y=2
x2y(dy/dx)+y^2+x^2(dy/dx)+2xy=0
(dy/dx)(2xy+x^2)=(-y^2-2xy)
dy/dx=(-y^2-2xy)/(2xy+x^2)
dy/dx=(-1-2)/(2+1)=-3/3= -1

So the slope of the tangent line at point (1,1) on that curve is -1.

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