The following are all problems I didn't assign from section 6.1 of the text. Here are the solutions done out. Use these as extra practice problems.
For exercises 39-42, the velocity \(\displaystyle v = \frac{ds}{dt}\) or acceleration \(\displaystyle a = \frac{dv}{dt}\) of a body moving along a coordinate line is given. Find the body's position \( s \) at time \(t\)
\( \displaystyle \frac{ds}{dt} = 9.8t + 5\) so \(\displaystyle ds = (9.8t + 5)dt. \) Then \(\displaystyle \int ds = \int (9.8t + 5)dt\) so \( \displaystyle s = 4.9t^2 + 5t + C.\) Using the initial condition: \( \displaystyle 10 = 4.9 \cdot 0^2 + 5 \cdot 0 + C \) \( \displaystyle \implies C = 10.\) Therefore, \( \displaystyle s = 4.9t^2 + 5t + 10.\)
\( \displaystyle \frac{ds}{dt} = \sin{(\pi t)},\) so \( ds = \sin{ (\pi t )} dt\) and \( \displaystyle \int ds = \int \sin{ (\pi t )} dt.\) Then \( \displaystyle s = \frac { - \cos {( \pi t)} } {\pi} + C.\) Using the initial condition, we have: \( \displaystyle 0 = \frac { - \cos {( \pi \cdot 1)} } {\pi} + C \implies C = - \frac{1}{\pi}.\) Answer: \( \displaystyle s = - \frac{\cos{( \pi t)}}{\pi} - \frac{1}{\pi}.\)
\(\displaystyle a = \frac{dv}{dt} = 32\) So \(\displaystyle dv = 32 dt, \) then \(\displaystyle \int dv = \int 32 dt \implies \) \(\displaystyle v = 32t + C.\) Since \( v(0) = 20,\) \( 20 = 32 \cdot 0 + C \implies \) \(C = 20.\) Then \(\displaystyle v = \frac{ds}{dt} = 32t + 20,\) so \( ds = (32t + 20)dt \implies \) \( \displaystyle \int ds = \int (32t + 20) dt.\) Then \(\displaystyle s = 16t^2 + 20t + C.\) Using the other initial condition that \( s(0) = 0,\) we have \( 0 = 16 \cdot 0^2 + 20 \cdot 0 + C \implies \) \(C = 0.\) Answer: \( s = 16t^2 + 20t.\)
\(\displaystyle a = \frac{dv}{dt} = \cos t. \) So \(\displaystyle dv = \cos t dt, \) then \(\displaystyle \int dv = \int \cos t dt \implies \) \(\displaystyle v = \sin t + C.\) Since \( v(0) = -1,\) \( -1 = \sin 0 + C \implies \) \(C = -1.\) Then \(\displaystyle v = \frac{ds}{dt} = \sin t - 1,\) so \( ds = (\sin t - 1)dt \implies \) \( \displaystyle \int ds = \int (\sin t - 1) dt.\) Then \(\displaystyle s = -\cos t - t + C.\) Using the other initial condition that \( s(0) = 1,\) we have \( 1 = - \cos 0 - 0 + C \implies \) \(C = 2.\) Answer: \( s = - \cos t - t + 2.\)
\( \displaystyle dr = (3x^2 - 6x + 12) dx \implies \) \( \displaystyle \int dr = \int (3x^2 - 6x + 12) dx \implies \implies \) \( \displaystyle r = x^3 - 3x^2 + 12x + C.\) Plugging in the initial condition, we have \( 0 = 0^3 - 3 \cdot 0^2 + 12 \cdot 0 + C \implies \) \( \displaystyle C = 0.\) The answer is \( \displaystyle r = x^3 - 3x^2 + 12x \).
\( \displaystyle dc = (3x^2 - 12x + 15) dx \implies \) \( \displaystyle \int dc = \int (3x^2 - 12x + 15) dx \implies \implies \) \( \displaystyle c = x^3 - 6x^2 + 15x + C.\) Plugging in the initial condition, we have \( 400 = 0^3 - 6 \cdot 0^2 + 15 \cdot 0 + C \implies \) \( \displaystyle C = 400.\) The answer is \( \displaystyle c = x^3 - 6x^2 + 15x + 400 \).
Start by simplifying the notation. Rewrite \( \displaystyle \frac{d^2s}{dt^2} = -5.2 \) as \( \displaystyle \frac{ds'}{dt} = -5.2\) then separate variables. \( \displaystyle ds' = -5.2dt \implies \) \( \displaystyle \int ds' = \int -5.2dt \implies \) \( \displaystyle s' = -5.2t + C.\) Use the initial condition that \( \displaystyle \frac{ds}{dt} = 0 \) when \( t = 0\) to get \( 0 = -5.2 \cdot 0 + C \implies \) \( \displaystyle C = 0.\) So \( \displaystyle s' = -5.2t.\) Now rewrite as a differential equation again, and separate variables: \( \displaystyle \frac{ds}{dt} = -5.2t \implies \) \( \displaystyle ds = -5.2dt \implies \) \( \displaystyle \int ds = -5.2 \int dt \implies \) \( \displaystyle s = -2.6t^2 + C.\) Use the second initial condition, that \( s = 4\) when \(t = 0\): \( 4 = -2.6 \cdot 0^2 + C \implies \) \( \displaystyle C = 4.\) So the height as a function of time is \(s = -2.6t^2 + 4.\) To find when the feather and hammer hit the moon, set \(s(t) = 0\) and solve for t: \( 0 = -2.6t^2 + 4 \implies \) \( \displaystyle t = 1.240\) seconds
Start by simplifying the notation. Rewrite \( \displaystyle \frac{d^2y}{dx^2} = 6x \) as \( \displaystyle \frac{dy'}{d6} = 6x\) then separate variables. \( \displaystyle dy' = 6x dx \implies \) \( \displaystyle \int dy' = \int 6x dx \implies \) \( \displaystyle y' = 3x^2 + C.\) Use the initial condition that the tanent line to the curve is horizontal at the point \( (0, 1) \). For that to be the case, \( y' = 0 \) when \( x = 0.\) So \( 0 = 3 \cdot 0^2 + C \implies \) \( \displaystyle C = 0.\) So \( \displaystyle y' = 3x^2.\) Now rewrite as a differential equation again, and separate variables: \( \displaystyle \frac{dy}{dx} = 3x^2 \implies \) \( \displaystyle dy = 3x^2 dx \implies \) \( \displaystyle \int dt = \int 3x^2 dx \implies \) \( \displaystyle y = x^3 + C.\) Use the second initial condition, that the curve passes through the point \( (0, 1)\): \( 1 = 0^3 + c \implies \) \( \displaystyle C = 1.\) So \(y = x^3 + 1.\) The solution to the initial value differential equation is unique.