Section 6.11 Antiderivatives and Differential Equations
These answers correspond to the problems in Section 3.6.
\(\ds -\frac{44}{3}+2e(1-6e^2)\text{.}\)
\(\ds g(x)=\ln|x|+\arctan x+c\text{.}\)
\(\ds f(t)=2e^t-3\sin t+t+2\text{.}\)
\(\ds f(x)=\frac{\sin 8x}{4}+\frac{1}{3}e^{3x}-x+\frac{2}{3}\text{.}\)
\(f(x)=Ae^{kx}\text{.}\)
\(\ds f(x)=2\sin x+2x^4-e^x+8\text{.}\)
\(\displaystyle \ds g(x)=-\cos x -x^{-1}-e^x +\pi ^{-1}+e^{\pi }\)
\(\ds f(x)=\frac{1}{3}x^3+x^2+3x-\frac{242}{3}\) and \(\ds f(1)=-\frac{229}{3}\text{.}\)
\(\ds h(1)=2e(1-e^2)-\frac{44}{3}\text{.}\)
\(\ds F(z)=\frac{1}{2}\ln (z^2+9)\text{.}\)
It is given that \(f(0)=1\) and \(f'(0)=0\text{.}\) Thus \(f(x)=x^3+1\text{.}\)
\(\ds \int \frac{dx}{x(1+\ln x)}=\ln(1+\ln x)+C\text{.}\)
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\(\displaystyle \ds F(x)=-\frac{1}{9}(1-x)^9\)
\(\displaystyle \ds \int \tan ^2xdx=\int (\sec ^2x-1)dx=\tan x-x+C\)
\(\displaystyle \ds F(x)=\frac{1}{\sqrt{2}}\arctan \frac{x}{\sqrt{2}}+C\)
\(\displaystyle \ds F(x)=\frac{1}{6}e^{3x}+\frac{1}{2}e^x+C\)
\(f(t)=2e^t-3\sin t+t-2\text{.}\)
\(\displaystyle s(t)=2e^t-3\sin t-(2e^{\pi}+3)t-2\)
\(s(t)=-\sin t-3\cos t+3t+3\text{.}\)
(a) \(v(t)=-10t+5\text{;}\) (b) \(s(t)=-5t^2+5t+30\text{;}\) (c) \(t=6\) s; (d) \(|v(6)|=5\) m/s.
It is given that \(x(0)=10\text{,}\) \(x'(0)=v(0)=0\) and \(x"(t)=12t\text{.}\) Hence \(x(t)=2t^3+10\text{.}\)
(a) \(\ds v(t)=\frac{3}{2}t^2+6\text{.}\) (b) 4 seconds. Solve \(v(t)=30\text{.}\)
(a) Let \(s(t)\) be the height of the ball after \(t\) seconds. It is given that \(s(0)=0\text{,}\) \(s'(0)=v(0)=64\) ft/sec and \(s"(0)=v'(0)=a(0)=-32\) ft/sec\(^2\text{.}\) Thus \(s(t)=-16t^2+64t=16t(4-t)\text{.}\) From \(s(4)=0\) it follows that the ball is in the air for 4 seconds. (b) \(v(4)=s'(4)=-64\) ft/sec\(^2\text{.}\)
(a) From the fact that the velocity of an falling object is approximated by \(v(t)=-gt+v(0)\) and the fact that, in the given case, \(v(t)=0\text{,}\) we conclude that the distance \(y=y(t)\) between the ball and the surface of the Earth at time \(t\) is given by \(\ds \frac{dy}{dt}=-gt\text{.}\) Hence, \(\ds y=-\frac{gt^2}{2}+H\text{,}\) where \(H\) is the height of the blimp at the moment when the ball was dropped. At the moment when the ball hits surface we have that \(\ds 0=-\frac{gt^2}{2}+H\) which implies that it takes \(\ds t=\sqrt{\frac{2H}{g}}\) seconds for a ball to drop \(H\) metres. (b) \(\ds v=-\sqrt{2Hg}=-10\cdot 7=-70\) m/sec.
\(\ds y=\frac{1}{3}\cdot(-2\cos 3x +x^3+e^{3x}+1)+x\text{.}\)
\(\ds y=\sin \left( x+\frac{\pi }{2}\right)\text{.}\)
\(\ds y=\tan \left( x-\frac{\pi }{4}\right)\text{.}\)
\(\ds y=4e^x-1\text{.}\)
\(\displaystyle \ds x(t)=-3(4t-7)^{-3}+4\)
\(\ds y=\ln (x+e^2)\text{.}\)
\(\ds y=2-e^{-t}\text{.}\)
(a) \(\ds y=\frac{3}{2}\sin 2x-\frac{1}{4}\exp (-4x)+\frac{5}{4}\text{.}\) (b) \(\ds F(x)=\frac{1}{2}\ln (2x+1)+C\text{.}\)
\(v(0)=80\) ft/s.
Let \(x\) be the number of towels sold per week at the price \(p=p(x)\text{.}\) Let \(C=C(x)\) be the cost of manufacturing \(x\) towels. It is given that \(\ds \frac{dC}{dx}=0.15\) CAD/towel and \(\ds \frac{dp}{dx}=-\frac{0.10}{50}\) CAD/towel. Hence \(C(x)=0.15x+a\) and \(\ds p(x)=-\frac{0.10x}{50}+b\text{,}\) for some constants \(a\) and \(b\) (in CAD). Then the profit is given by \(\ds P=P(x)=\mbox{Revenue} -\mbox{Cost} =x\cdot p(x)-C(x)=-\frac{0.10x^2}{50}+bx-0.15x-a\text{.}\) The quantity that maximizes revenue is \(x=1000\) towels and it must be a solution of the equation \(\ds \frac{dP}{dx}=-\frac{0.10x}{25}+b-0.15=0\text{.}\) Hence \(\ds -\frac{0.10\cdot 1000}{25}+b-0.15=0\) and \(b=4.15\) CAD. The price that maximizes the profit is \(\ds p=-\frac{0.10\cdot 1000}{50}+4.15=2.15\) CAD.