We covered the concept of differential calculus in just the last post. It is time to slow down just a bit, catch our breath, and explore more in depth what this all means.
First, we started out with the concept of a function. We used the example of a curve in space where the x-coordinate (and all of the others) depend on an index “t” that indicates where each point is along this curve in space.
x(t) = x-coordinate of a point at the index t
If you specify a “t” value, the function x(t) will give the x-coordinate of a point along the curve that corresponds to “t”.
It might not have been perfectly clear, but we assumed that the curve is made out of points that are infinitely close to each other (same as a line). There are no gaps or kinks in the curve, such that when you increase “t” by an infinitesimally small amount “dt” it will return the coordinates of a point that is infinitesimally close along the curve. This is the concept of a “smooth” or “continuous” function.
An example of this is a car traveling down a straight road. Suppose the function x(t) shows how far the car has gone down the road at a time “t”. At each time, “t”, the function x(t) will tell you the exact position of the car. Now since a car can’t instantaneously disappear from one point and appear in a new point, it always takes time to travel continuously between any two points, which means that if you look at the position of the car a billionth of a second later, the position will be very close to where it was before.
As we discussed, “dt” is much smaller than a billionth of a second, it is the smallest amount possible without being zero. If “t” has increased by a small amount, “t+dt”, then the x-coordinate also will have changed by a very small amount “dx”. This is what we mean by the expression:
dx = x(t+dt) – x(t)
The derivative dx/dt is defined to be:
dx/dt = [x(t+dt) – x(t)]/dt
which indicates how fast “x” is changing in relation to a change in “t” at the point “t”. We can now hopefully see why differential calculus is the mathematics of motion. In the example of the car above, dx/dt is velocity of the car at any instant “t” along its path – it moves a distance dx in the time dt.
In the example where:
x(t) = t2
x(t+dt) = (t+dt)2
dx = x(t+dt) – x(t)
= (t+dt)2 – t2
= t2+2*t*dt + dt2 – t2 (the t2 cancels out)
= 2*t*dt + dt2
and
dx/dt = [x(t+dt) – x(t)]/dt
= 2*t (when dt –> zero)
We say that the derivative of t2 is 2*t, which also indicates how fast x is changing in ratio to the change in t at every point along the curve.
This function and differential notation is very powerful, we can get the answers to some questions without even knowing the exact function.
Suppose that x(t) is the sum of two functions of t, f(t), g(t):
x(t) = f(t) + g(t) where f(t) and g(t) are smooth
df = f(t+dt) – f(t)
dg = g(t+dt) – g(t)
dx = x(t+dt) – x(t)
= f(t+dt) + g(t+dt) – (f(t) + g(t))
= [f(t+dt) – f(t)] + [g(t+dt) – g(t)]
= df + dg
Thus the derivative is also the sum of the derivatives:
dx/dt = df/dt + dg/dt.
Suppose that x(t) is the product of two functions of t:
x(t) = f(t)*g(t) where f(t) and g(t) are smooth
dx = x(t+dt) – x(t)
= f(t+dt)*g(t+dt) – f(t)*g(t)
=[df + f(t)]*[dg + g(t)] – f(t)*g(t)
= df*dg + df*g(t) + f(t)*dg + f(t)*g(t) – f(t)*g(t)
= df*dg + df*g(t) + f(t)*dt
Thus the derivative is:
dx/dt = df*(dg/dt) + (df/dt)*g(t) + f(t)*(dg/dt)
= (df/dt)*g(t) + f(t)*(dg/dt) because df –> 0
dx/dt = (df/dt)*g(t) + f(t)*(dg/dt)
Notice that we must always divide differentials by differentials to get derivatives, and all differentials standing alone (dx, df, dg, dt) go to zero.
We can now use the product rule to get the derivative of:
x(t) = t3 = t2 * t {f(t) = t2, g(t) = t}
we already know that df/dt = 2*t from above and
dg/dt = [t+dt – t]/dt = 1
therefore since
dx/dt = (df/dt)*g(t) + f(t)*(dg/dt)
dx/dt = (2*t)*t + t2*1
= 3*t2
By extending this we can show that for any power of t:
If x(t) = tn = t(n-1) * t
then
dx/dt = n * t(n-1)
We can see where some of the derivative tables in your calculus book come from.
We will end off by finding the derivative for a ratio of functions:
x(t) = f(t)/g(t)
dx/dt = [f(t+dt)/g(t+dt) – f(t)/g(t)]/dt
= [(df+f(t))/(dg+g(t)) – f(t)/g(t)]/dt
= [[g(t)*(df+f(t))-f(t)*(dg+g(t))]/(g(t)*(dg+g(t))]/dt
= [g(t)*(df/dt) + f(t)*g(t)/dt – f(t)*(dg/dt) – f(t)*g(t)/dt]/ (g(t)*dg + g(t)*g(t))
and when df and dg –> 0,
dx/dt = [g(t)*(df/dt) – f(t)*(dg/dt)]/(g(t)*g(t))
When we learn the “algebra” of differentials, we can find derivatives for a vast array of smooth functions.
As we have seen, when we know the derivatives of a curve, we can find the direction of the tangent line at any point along the curve. Recall from the last post in 2-D:
(Dx, Dy) = (dx/dt, dy/dt)/((dx/dt)2+(dy/dt)2)1/2
The nice thing about this notation is that it is easy to go to 3-D, 4-D, and as many -D’s that you want just by continuing the pattern.
