BC Calculus : §6.3 Extra Practice Problems on Differential Equations

• Class: 5H: BC Calculus
• Author: Peter Atlas
• Text: Calculus Finney, Demana, Waits, Kennedy

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 $$t$$

1. $$\displaystyle v = 9.8t + 5; \text{ } s(0) = 10$$
Solution

$$\displaystyle \frac{ds}{dt} = 9.8t + 5$$ 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, $$\displaystyle s = 4.9t^2 + 5t + 10.$$

2. $$\displaystyle v = \sin{ (\pi t)}; \text{ } s(1) = 0$$
Solution

$$\displaystyle \frac{ds}{dt} = \sin{(\pi t)},$$ 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: $$\displaystyle s = - \frac{\cos{( \pi t)}}{\pi} - \frac{1}{\pi}.$$

3. $$\displaystyle a = 32; \text{ } s(0) = 0; \text{ } v(0) = 20$$
Solution

$$\displaystyle a = \frac{dv}{dt} = 32$$ 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: $$s = 16t^2 + 20t.$$

4. $$\displaystyle a = \cos t; \text{ } s(0) = 1; \text{ } v(0) = -1$$
Solution

$$\displaystyle a = \frac{dv}{dt} = \cos t.$$ 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: $$s = - \cos t - t + 2.$$

1. Suppose that the marginal revenue when $$x$$ thousand units are sold is $$\displaystyle \frac{dr}{dx} = 3x^2 - 6x + 12$$ dollars per unit. Find the revenue function $$r(x)$$ if $$r(0) = 0.$$
Solution

$$\displaystyle dr = (3x^2 - 6x + 12) dx \implies$$ $$\displaystyle \int dr = \int (3x^2 - 6x + 12) dx \implies \implies$$ $$\displaystyle r = x^3 - 3x^2 + 12x + C.$$ 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 $$\displaystyle r = x^3 - 3x^2 + 12x$$.

1. Suppose that the marginal cost of manufacturing an item when $$x$$ thousand units are produced is $$\displaystyle \frac{dc}{dx} = 3x^2 - 12x + 15$$ dollars per item. Find the cost function $$c(x)$$ if $$c(0) = 400.$$
Solution

$$\displaystyle dc = (3x^2 - 12x + 15) dx \implies$$ $$\displaystyle \int dc = \int (3x^2 - 12x + 15) dx \implies \implies$$ $$\displaystyle c = x^3 - 6x^2 + 15x + C.$$ 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 $$\displaystyle c = x^3 - 6x^2 + 15x + 400$$.

1. When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about 4 ft above the ground. The television footage of the event shows the hammer and feather falling more slowly than on Earth, where, in a vacuum, they would have taken only half a second to fall the 4 feet. How long did it take the hammer and feather to fall 4 ft on the moon? To find out, solve the following initial value problem for s as a function of t. Then find the value of t that makes s equal to 0. Differential equation: $$\displaystyle \frac{d^2s}{dt^2} = -5.2$$ ft/sec2; Initial conditions: $$\displaystyle \frac{ds}{dt} = 0$$ and $$s = 4$$ when $$t = 0.$$
Solution

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

1. Let $$\displaystyle \frac{d^2y}{dx^2} = 6x.$$ Find a solution to the differential equation that is continuous for $$- \infty \lt x \lt \infty$$ and whose graph passes through the point $$(0,1)$$ and has a horizontal tangent there.
Solution

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.