# Antiderivatives

## 4.7.48

• A boat leaves a dock at 2:00pm, and travels due South at a speed of 20 km/h. Another boat has been heading due East at 15 km/h, and reaches the same dock at 3:00pm. At what time were the 2 boats closest together?
• The 4 variables are the coordinates of the 2 boats: we have $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$. We can choose the dock as the origin $$O(0,0)$$.
• The function to minimize is the distance $$d(A,B)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$.
• We'll be able to write all 4 coordinates in terms of the time variable $$t$$, which is measured in hours past 2:00pm.
• Boat A is travelling South at 20 km/h, therefore we have $$x_1'=0$$ and $$y_1'=20$$.
• Boat B is travelling East at 15 km/h, therefore we have $$x_2'=15$$ and $$y_2'=0$$.
• Since boat A was at the dock at 2:00pm, we have $$x_1(t=0)=0$$ and $$y_1(t=0)=0$$.
• Since boat B reached the dock at 3:00pm by travelling 15 km/h due East, it was 15 km East of the dock at 2:00pm. That is, we have $$x_2(t=0)=-15$$ and $$y_2(t=0)=0$$.

## 4.7.48 continued

• Now we can find the coordinate functions.
• The function $$x_1$$ satisfies $$x_1'=0$$. Therefore, it is a constant function: $$x_1(t)=c_1$$.
• To find the constant $$c_1$$, we can substitute the initial value: $$x_1(t=0)=c_1=0$$, therefore $$x_1(t)=0$$.
• Similarly, $$y_1$$ will be a function such that $$y_1'=20$$.
• By the differentiation rules, you can see that $$y_1=20t+d_1$$.
• Substituting the initial value (IV) $$y_1(t=0)=d_1=0$$, we get $$y_1(t)=20t$$.
• Similarly, we can get $$x_2(t)=15t-15$$ and $$y_2(t)=0$$.
• Finally, we can make the function to minimize a single variable function, and find its absolute minimum.
• We have $$d(A,B)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(15t-15)^2+(20t)^2}=5\sqrt{25t^2-18t+9}$$.
• This gives $$d'=\frac52\frac{50t-18}{\sqrt{25t^2-18t+9}}$$.
• Then we can see that the absolute minimum is attained at $$t=\frac{9}{25}$$.

## Antiderivatives

• We have seen that an important part of solving the problem was to find the general form of a function with a fixed derivative.
• Definition. Let $$f$$ be a function on an open interval $$I$$. We say that another function $$F$$ is an antiderivative of $$f$$, if we have $$F'(x)=f(x)$$ for all $$x$$ in $$I$$.
• Note that $$f$$ does not need to be differentiable. For example, you can check that the function $F(x)=\begin{cases} x^2 & 0\le x \\ -x^2 & x\le 0 \end{cases}$ is differentiable, and its derivative is $$f(x)=2|x|$$. Therefore, $$F$$ is an antiderivative of the function $$f$$, where the latter is not differentiable at $$x=0$$.
• Let's suppose that $$F_1$$ and $$F_2$$ are both antiderivatives of $$f$$. Then we have $F_1'=f=F_2'.$
• Therefore, we have $$(F_1-F_2)'=0$$ on the interval $$I$$. That is, $$F_1-F_2=C$$, a constant.
• Theorem. If $$F$$ is an antiderivative of $$f$$ on an interval $$I$$, then the most general antiderivative of $$f$$ on $$I$$ is $$F(x)+C$$, where $$C$$ is an arbitrary constant.

## Examples

• Find the general antiderivatives of the following functions.
• $$f(x)=\sin(x)$$.
• Since we know that $$(\cos x)'=-\sin x$$, we get $$(-\cos x)'=\sin x$$. Therefore, $$F(x)=-\cos x$$ is an antiderivative of $$f(x)=\sin x$$. This shows that the general antiderivative of $$f$$ is $$F(x)=-\cos x+C$$, where $$C$$ is an arbitrary constant.
• $$f(x)=x^{-1}$$.
• We have seen that $$(\ln|x|)'=x^{-1}$$. Therefore, the general antiderivative is $$F(x)=\ln|x|+C$$.
• $$f(x)=x^n,\,n\ne-1$$.
• We have seen that $$(x^{n+1})'=(n+1)x^n$$. This gives $$\left(\frac{x^{n+1}}{n+1}\right)'=x^n$$. Therefore, the general antiderivative of $$f$$ is $$F(x)=\frac{x^{n+1}}{n+1}+C$$.
• Exercises. 4.9: 2, 6, 10, 16

## Differential equations and rectilinear motion

• Differential equations.
• Note that an antiderivative $$F$$ is a solution of the equation $$F'=f$$.
• Equations in which the function to find is differentiated are called differential equations (DE).
• The general solution is the general antiderivative $$F(x)+C$$.
• Often, we get an additional condition called initial condition (IC) of the form $$F(x=x_0)=F_0$$.
• A differential equation together with an initial condition is called an initial value problem (IVP).
• Rectilinear motion
• Provided we can find antiderivatives, we can now solve problems about rectilinear motion, that is motion on a straight line.

## 4.9.65

• A stone is dropped from the upper observation deck of the CN Tower, 450 m above the ground. We will want to study its height $$h(t)$$. By gravity, its acceleration is $$a(t)=-10$$ m/s$${}^2$$.
• Its velocity $$v(t)$$ will satisfy $$v'(t)=a(t)$$. Since the problem doesn't say anything about initial speed, we'll assume that there wasn't any: $$v(t=0)=0$$.
• Let's solve this IVP. The general solution of the DE is $$v(t)=-10t+v_0$$. Then the IC gives $$v(t=0)=v_0=0$$, therefore we get $$v(t)=-10t$$.
• The height will satisfy another IVP: $$h'(t)=v(t),\,h(t=0)=450$$. The general solution is $$h(t)=-5t^2+h_0$$, therefore we get $$h(t)=-5t^2+450$$.
• Let's see how long does it takes the stone to reach the ground. We need $$-5t^2+450=0$$. Since we need $$t\ge0$$, the solution is $$t=3\sqrt{10}\approx9.4$$ s.
• The stone strikes the ground with the speed $$v(t=3\sqrt{10})=3\sqrt{10}10$$ m/s, which is about 342 km/h.
• If instead we had thrown the stone downward with $$5$$ m/s, then its speed would be $$v(t)=-10t+v_0=-10t-5$$, and thus its height would be $$h(t)=-5t^2-5t+450$$. In this case, it would reach the ground at $$t=\frac{-1+\sqrt{361}}{2}=9$$.
• Exercises. 4.9: 38, 74, 76