How do you use differentials and the function #f(x,y) = arctan(x*y^2)# to approximate the value of f(0.94, 1.17)?

Answer 1

#L(x,y)=0.92539816#. See below.

If we stay near the point of tangency (#x_0,y_0#), then the tangent plane serves as a linear approximation of #f(x,y)#.

The tangent plane is given by:

#z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)#

And so we have:

#z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)=L(x,y)#
Where #L(x,y)# is the linear approximation of #f(x,y)#. Essentially, when #(x,y)~~(x_0,y_0)#, then #f(x,y)~~L(x,y)#.
Given that #f(x,y)=arctan(x*y^2)#, we can use linear approximation to estimate #f(0.94,1.17)#.
First, we pick "nice" points for the approximation which are near the original #x# and #y# values. Integers are great choices. Given #x=0.94# and #y=1.17#, I would use #x_0=1# and #y=1#. These are much more pleasant to work with.
#f(x,y)=arctan(x*y^2)#
We take the partial derivative of the function with respect to #x#:
#(del)/(delx)=y^2/(x^2y^4+1)#
Then with respect to #y#:
#(del)/(dely)=(2xy)/(x^2y^4+1)#
Now, just as we do when finding the tangent plane, we calculate #f_x(x_0,y_0)#, #f_y(x_0,y_0)#, and #f(x_0,y_0)#. From above, we've chosen #(x_0,y_0)=(1,1)#.
#f_x(1,1)=(1^2)/(2)=0.5#
#f_y(1,1)=2/2=1#
#f(1,1)=arctan(1)=pi/4#

Revisiting our equation for the tangent plane:

#z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)#

Putting in our values:

#z=pi/4+0.5(x-1)+1(y-1)#
#=>z=pi/4+x/2-1/2+y-1#
#=>z=x/2+y+pi/4-3/2#
Where #z=L(x,y)=>L(x,y)=x/2+y+pi/4-3/2#.
This means that when #(x,y)# is close to #(1,1)#, #arctan(x*y^2)~~x/2+y+pi/4-3/2#.
Finally, our linear approximation is found by putting the actual #x# and #y# values into the equation we found for #L(x,y)#:
#f(0.94,1.17)~~(0.94/2)+1.17+pi/4-3/2=0.92539816#
We can check the accuracy of our approximation by comparing it to the actual value of #arctan(x*y^2)# for #(0.94,1.17)#.
#f(0.94,1.17)=arctan((0.94)*(1.17)^2)#
#f(0.94,1.17)=0.910149371#
This is quite close to our approximation, with a difference of about #1.7%#.
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Answer 2

To approximate the value of ( f(0.94, 1.17) ) using differentials, we'll use the total differential formula:

[ df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy ]

First, we find the partial derivatives of ( f(x, y) = \arctan(xy^2) ) with respect to ( x ) and ( y ):

[ \frac{\partial f}{\partial x} = \frac{y^2}{1 + (xy^2)^2} ] [ \frac{\partial f}{\partial y} = \frac{2xy}{1 + (xy^2)^2} ]

Then, we'll evaluate these partial derivatives at the point ( (0.94, 1.17) ) and compute ( dx ) and ( dy ) by taking the differences between the given point and the nearby point ( (x, y) ):

[ dx = 0.94 - 0.94 = 0 ] [ dy = 1.17 - 1.17 = 0 ]

Now, we'll substitute the values into the formula to find the approximate value of ( df ):

[ df = \frac{1.17^2}{1 + (0.94 \times 1.17^2)^2} \times 0 + \frac{2 \times 0.94 \times 1.17}{1 + (0.94 \times 1.17^2)^2} \times 0 ]

[ df = 0 ]

Therefore, the approximate value of ( f(0.94, 1.17) ) using differentials is ( 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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