Implicit differentation

crouchjr

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Oct 30, 2009
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Anyone know how to do this problem?
Use implicit differentiation to find the slope of the tangent line to the curve

-2x^2-1xy-4y^3=-110

at the point (-2,3)

What does m=?
 
Step #1 Make sure the point is on the curve. Please demonstrate this.

Step #2 Use your Implicit Differentiation skills. This requires a nice understanding fo the Chain Rule

For the first term, I get -4x <== Just the good old polynomial rule there.

For the second term, I get -x*(dy/dx) - y <== A thing of beauty. Product rule and chain rule.

You do the third term and the other side of the equal sign.

Step #3 Find dy/dx

Step #4 Evaluate dy/dx at the given point (assuming it is on the curve).
 
Close. How about 12y2dydx\displaystyle -12y^{2} \cdot \frac{dy}{dx}. You missed th esign and the exponent.

Great. Now put it all together and find an expression for dy/dx.

Wait! Did you check to see that the point IS on the curve?
 
Given: 2x2xy4y3+110 = 0, and point(2,3)\displaystyle Given: \ -2x^{2}-xy-4y^{3}+110 \ = \ 0, \ and \ point(-2,3)

Find line tangent to f(x,y) at the point.\displaystyle Find \ line \ tangent \ to \ f(x,y) \ at \ the \ point.

Implicit differentiation yields: 4x[y+xy]12y2y+0 = 0\displaystyle Implicit \ differentiation \ yields: \ -4x-[y+xy']-12y^{2}y'+0 \ = \ 0

4xyxy12y2y = 0      y = 4x+yx+12y2, at (2,3), m = 5106.\displaystyle -4x-y-xy'-12y^{2}y' \ = \ 0 \ \implies \ y' \ = \ -\frac{4x+y}{x+12y^{2}}, \ at \ (-2,3), \ m \ = \ \frac{5}{106}.

Hence, y3 = 5106(x+2), y = 5x+328106.\displaystyle Hence, \ y-3 \ = \ \frac{5}{106}(x+2), \ y \ = \ \frac{5x+328}{106}.

See graph.\displaystyle See \ graph.

[attachment=0:1gseb7b0]zero.jpg[/attachment:1gseb7b0]
 

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