Question

Consider the curve y=4x-x^3 and the chord AB joining points A(-3, 15) and B(2, -15) on the curve. Find the x- and y-coordinates of the points on the curve where the tangent line is parallel to chord AB.

Answer 1

We are asked to find the coordinates of the points where the slope of the curve is equal to the slope of the chord AB. We are given the coordinates of A and B. We can use these and the slope formula to find the slope of the chord.

Recall the slope formula is: $m= \frac{ y_{2} - y_{1} }{ x_{2} - x_{1} }$
We label the points as follows (note that you could change the way you label the points as you will get the same slope regardless of what point is designated with the 1s and which is designated with the 2s).
$A=( x_{1} , y_{1)}=(-3,15)$
$B=( x_{2} , y_{2}=(2,-15)$
Plugging these into the slope formula yields: $m= \frac{-15-15}{2--3 }= \frac{-30}{5} =-6$
The slope of the chord is -6


Next consider the curve $y=4x- x^{3}$. We can find the slope of the curve by taking the first derivative. Thus, the slope of the curve is given by: $y^{'} =4-3 x^{2}$

Let us set the slope of the curve equal to -6 to find the points (x,y) where the chord and the curve have the same slope.

That is, $-6=4-3 x^{2}$
$3 x^{2} =10$
$x^{2} = \sqrt{ \frac{10}{3} }$
$x= \sqrt{ \frac{10}{3} },x= -\sqrt{ \frac{10}{3} }$

We now have the x-coordinates of the points at which the slope of the chord equals that of the curve. To find the y-coordinates we substitute the values we found for x in the original equation as follows:

$y=4( \sqrt{ \frac{10}{3} }) -( \sqrt{ \frac{10}{3} }) ^{3} =4( \frac{10}{3}) ^{ \frac{1}{2} }-( \frac{10}{3}) ^{ \frac{3}{2} }$
and
$y=4( -\sqrt{ \frac{10}{3} }) -(- \sqrt{ \frac{10}{3} }) ^{3} =-4( \frac{10}{3}) ^{ \frac{1}{2} }+( \frac{10}{3}) ^{ \frac{3}{2} }$

Thus the two points we seek (x,y) are:
$( \sqrt{ \frac{10}{3} }, 4( \frac{10}{3}) ^{ \frac{1}{2} }-( \frac{10}{3}) ^{ \frac{3}{2} } )$
and
$( -\sqrt{ \frac{10}{3} }, -4( \frac{10}{3}) ^{ \frac{1}{2} }+( \frac{10}{3}) ^{ \frac{3}{2} } )$