Friday, May 22, 2015

Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 42

Suppose that the curve has equation of x+y=c. Show that the sum of the x and y intercepts of any tangent line to the curve is equal to c.

Taking the derivative of the curve implicitly we have.


12x+dydx2y=0dydx=yx


Using Point Slope Form


yy1=m(xx1)yy1=y1x1(xx1)


Recall that x-intercept is the value of x where y=0 consequently, y-intercept is the value of y where x=0. So,


For x=0,yy1=y1x1(0x1)y=y1y1x1(x1)y=y1+y1x1For y=0,0y1=y1x1(xx1)xx1=y1x1y1x=y1x1y1+x1x=x1+x1y1


Thus, the x and y-intercepts are x=x1+x1y1 and y=y1x1y1

Now, solving for the condition...


(x1+x1y1)+(y1x1y1)=cx1+2x1y1+y1=cbut c=x+yc=(x+y)2c=x+xy+xy+yc=x+2xy+y


Therefore, it shows that the sum of the x and y-intercepts of any tangent line to the curve is equal to c.

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