Question
(a) Topic-8.4- Finding the Area Between Curves Expressed as Functions of \(x\)
(b) Topic-5.3- Determining Intervals on Which a Function is Increasing or Decreasing
(c) Topic-8.7- Volumes with Cross Sections: Squares and Rectangles
(d) Topic-3.1- The Chain Rule
2. Let f and g be the functions defined by f( x) = ln(x + 3) and \(g(x)=x^{4}+2x^{3}\) . The graphs of f and g, shown in the figure above, intersect at x = −2 and x = B, where B > 0.
(a) Find the area of the region enclosed by the graphs of f and g.
(b) For \(-2\leq x\leq B\), let h (x) be the vertical distance between the graphs of f and g. Is h increasing or decreasing at x = −0.5 ? Give a reason for your answer.
(c) The region enclosed by the graphs of f and g is the base of a solid. Cross sections of the solid taken perpendicular to the x-axis are squares. Find the volume of the solid.
(d) A vertical line in the xy-plane travels from left to right along the base of the solid described in part (c). The vertical line is moving at a constant rate of 7 units per second. Find the rate of change of the area of the cross section above the vertical line with respect to time when the vertical line is at position x = −0.5
▶️Answer/Explanation
2(a) Find the area of the region enclosed by the graphs of f and g.
\(In(x+3)=x^{4}+2x^{3}\Rightarrow x=-2,x=B=0.781975\)
\(\int_{-2}^{B}(f(x)-g(x))dx =3.603548\)
The area of the region is 3.604 (or 3.603 ).
2(b) For −2 ≤ x ≤ B , let h (x ) be the vertical distance between the graphs of f and g. Is h increasing or decreasing at x = −0.5 ? Give a reason for your answer.
h (x ) = f ( x) − g (x )
h′(x ) = f ′( x) − g′( x)
h′(−0.5) = f ′(−0.5) − g′(−0.5) = −0.6 (or −0.599 )
Since h′(−0.5) < 0, h is decreasing at x = −0.5.
2(c) The region enclosed by the graphs of f and g is the base of a solid. Cross sections of the solid taken perpendicular to the x -axis are squares. Find the volume of the solid.
\(\int_{-2}^{B}(f(x)-g(x))^{2}dx =5.340102\)
The volume of the solid is 5.340.
2(d) A vertical line in the xy -plane travels from left to right along the base of the solid described in part (c).The vertical line is moving at a constant rate of 7 units per second. Find the rate of change of the area of the cross section above the vertical line with respect to time when the vertical line is at position x = −0.5.
The cross section has area \(A(x)=(f(x)-g(x))^{2}\).
\(\frac{d}{dt}\left [ A(x) \right ]=\frac{dA}{dx}.\frac{dx}{dt}\)
\(\frac{d}{dt}\left [ A(x) \right ]|_{x=-0.5}=A'(-0.5).7 = -9.271842\)
At x = −0.5, the area of the cross section above the line is changing at a rate of −9.272 (or −9.271) square units per second.