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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

If the tangent at a point P on the curve y = x 3 intersects the curve again at Q , let A be the area of the region bounded by the curve and the line segment PQ. Let B be the area of the region defined in the same way starting with Q instead of P. What is the relationship between A and B?

To determine

To find:

The relationship between A and B

Explanation

1) Concept:

Area between curve:

The area A of the region bounded by curves y=fx, y=gx and the line x=c, x=d,  where f and g are continuous and f(x)g(x) for all x in [c, d], is

A=cdfx-gxdx

2) Calculation:

Assume P lies on the curve and in first quadrant.

As y=x3 lies in the first and third quadrant, this assumption is valid.

Let P=a, a3.

Use the below information to find the slope of the tangent line to curve y=x3 at the point P=(a, a3)

Here, fx=x3 and fa=a3, so the slope is

m=limxafx-fax-a=limxax3-a3x-a

=limxa(x-a)(x2+xa+a2)x-a

=limxa(x2+xa+a2)=3a2

Using point slope form, find the equation of tangent line at a, a3.

y-y1=m(x-x1)

y-a3=3a2(x-a)

y=3a2x-2a3

Now plot the curve and tangent line to find the point Q.

It is the intersection of the curve y= x3 and tangent line y=3a2x-2a3.

To find the point Q:

x3=3a2x-2a3

x3-3a2x+2a3=0

x-a2x+2a=0

So x=a  and x= -2a

Our point P corresponds to the first value x=a

So required x=-2a, y=-2a3=-8a3

So point Q is (-2a, -8a3).

Now we need to find the slope of the tangent line to the given curve passing through Q.

Here, fx=x3 and fa=-8a3, so the slope is

m=limxafx-fax-a=limx-2ax3+8a3x+2a

=limx-2a(x+2a)(x2-2xa+4a2)x+2a

=limx-2a(x2-2xa+4a2)=12a2

Using the point slope form, find the equation of the tangent line at (-2a,-8a3).

y-y1=m(x-x1)

y+8a3=12a2(x+2a)

y=12a2x+16a3

To find the point P2, which is the intersection of this tangent line at Q and the curve y=x3.

x3=12a2x+16a3

x3-12a2x-16a3=0

x+2a2x-4a=0

So, x=4a, y=4a3=64a3

So point P2 is (4a, 64a3)

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