• 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 \(t=\frac{9}{25}\).


  • 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.


  • 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.


  • 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