Exam 8: Integrals and Transcendental Functions

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Determine if the given function y = f(x) is a solution of the accompanying differential equation. -Differential equation: 2xy+2y=cosx2 x y ^ { \prime } + 2 y = \cos x Initial condition: y(π)=0y ( \pi ) = 0 Solution candidate: y=2sinxx\mathrm { y } = \frac { 2 \sin \mathrm { x } } { \mathrm { x } }

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A value of sinhx\sinh x or coshx\cosh x is given. Use the definitions and the identity cosh2xsinh2x=1\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1 to find the value of the other indicated hyperbolic function. - sinhx=43,tanhx=\sinh x = - \frac { 4 } { 3 } , \tanh x =

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Solve the problem. -A loaf of bread is removed from an oven at 350F350 ^ { \circ } \mathrm { F } and cooled in a room whose temperature is 70F70 ^ { \circ } \mathrm { F } . If the bread cools to 210F210 ^ { \circ } \mathrm { F } in 20 minutes, how much longer will it take the bread to cool to 185F185 ^ { \circ } \mathrm { F } .

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C

Verify the integration formula. - tanh1xdx=xtanh1x+12ln(1x2)+C\int \tanh ^ { - 1 } x d x = x \tanh ^ { - 1 } x + \frac { 1 } { 2 } \ln \left( 1 - x ^ { 2 } \right) + C

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Verify the integration formula. - 5xtanhx2dx=52ln(coshx2)+C\int 5 x \tanh x ^ { 2 } d x = \frac { 5 } { 2 } \ln \left( \cosh x ^ { 2 } \right) + C

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Find the slowest growing and the fastest growing functions as x→∞ . - y= y= y= y=

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Find the slowest growing and the fastest growing functions as x→∞ . - y=2x y=5x y= y=

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Solve the initial value problem. - d2ydx2=3ex,y(0)=1,y(0)=0\frac { \mathrm { d } ^ { 2 } y } { d x ^ { 2 } } = 3 e ^ { - x } , y ( 0 ) = 1 , y ^ { \prime } ( 0 ) = 0

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Solve the problem. -A certain radioactive isotope decays at a rate of 1%1 \% per 400 years. If tt represents time in years and y represents the amount of the isotope left, use the condition that y=0.99y0\mathrm { y } = 0.99 \mathrm { y } 0 to find the value of k\mathrm { k } in the equation y=y0ekt\mathrm { y } = \mathrm { y } _ { 0 } \mathrm { e } ^ { \mathrm { kt } } .

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Solve the differential equation. - dydx=9xy\frac { d y } { d x } = 9 \sqrt { x y }

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Find the derivative of y with respect to the appropriate variable. - y=(44θ)tanh1θ\mathrm { y } = ( 4 - 4 \theta ) \tanh ^ { - 1 } \theta

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Evaluate the integral. - 10sinh(4xln2)dx\int 10 \sinh ( 4 x - \ln 2 ) d x

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Solve the problem. -The velocity of a body of mass mm falling from rest under the action of gravity is given by the equation v=mgktanh(gkmt)\mathrm { v } = \sqrt { \frac { \mathrm { mg } } { \mathrm { k } } } \tanh \left( \sqrt { \frac { \mathrm { gk } } { \mathrm { m } } \mathrm { t } } \right) , where k\mathrm { k } is a constant that depends on the body's aerodynamic properties and the density of the air, gg is the gravitational constant, and tt is the number of seconds into the fall. Find the limiting velocity, limt\lim _ { \mathrm { t } \rightarrow \infty } , of a 210 lb. skydiver (mg=210)( \mathrm { mg } = 210 ) when k=0.006\mathrm { k } = 0.006 .

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Evaluate the integral. - 7π/67π/34cott7dt\int _ { 7 \pi / 6 } ^ { 7 \pi / 3 } 4 \cot \frac { t } { 7 } d t

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Rewrite the expression in terms of exponentials and simplify the results. - 6cosh(lnx)6 \cosh ( \ln x )

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Evaluate the integral. - x3ex4dx\int x ^ { 3 } e ^ { - x ^ { 4 } } d x

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Verify the integration formula. - 3csch3xdx=lntanh32x+C\int 3 \operatorname { csch } 3 x d x = \ln \left| \tanh \frac { 3 } { 2 } x \right| + C

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Solve the problem. -  Find the length of the curve x=y2324ln(y5),8y16\text { Find the length of the curve } x=\frac{y^{2}}{32}-4 \ln \left(\frac{y}{5}\right), 8 \leq y \leq 16 \text {. }

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Evaluate the integral. - 1e53tdt\int _ { 1 } ^ { \mathrm { e } ^ { 5 } } \frac { 3 } { \mathrm { t } } \mathrm { dt }

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Evaluate the integral. - 05π/4tanx5dx\int _ { 0 } ^ { 5 \pi / 4 } \tan \frac { x } { 5 } d x

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