The Calculus Diaries - Jennifer Ouellette [108]
But we want to find the exact value for the slope of the tangent line. The good news is that we have another nifty for mula for just this sort of problem: where Δ stands for a tiny increment. The bad news is that we have no idea what the values are for h2, h1, or Δh.
First things first: We need to find the value of h2. We can find h2 by plugging t2 into our starting function, like this: h(t2) = −16t22 + 200.
We know that t2 = t1 + Δt, so we can substitute that expression for t2, like this: h(t2 ) = −16(t1 + Δt)2 + 200.
We want to break this down as much as possible. Think of it as deconstructing our equation, i.e., reducing it to its individual components, the better to manipulate the pieces. For instance, we can rewrite the equation above as
h(t2) = −16(t1 + Δt ) (t1 + Δt ) + 200.
I’ll skip over the next few steps, which just involve further deconstruction according to the “grammatical rules” of math; suffice it to say, we end up with this:
h(t2)= −16(t12+2t1Δt +(Δt)2)+200.
Things are starting to get very confusing, and we’re not done yet. Next we need to find the value of h1. Happily, this is just our starting function: h1 = −16t12 +200.
Now we can subtract h1 from h2 to get Δh:
Δh= −16t12−32t1Δt−16(Δt)2+200−(−16t12+200)
Once we’re done deconstructing that equation and canceling out all the extraneous stuff, we end up with a far more malleable version: Δh =−32t1Δt−16(Δt)2.
Finally, we divide the whole mess by
This can be simplified even further to give us
Now our old friend the limit comes into play. We take the limit by setting Δt to .
We’ve already solved for above, so we can plug that value in and rewrite this as (-32t1 - 16∆t) = -32t.
When all is said and done, we end up with v(t) = −32t. Physics fans will recognize this as the stock formula for determining velocity: Velocity is equivalent to acceleration multiplied by time, written generically as v = at.
Derivatives: The Easy Way. I hope you feel a few twinges of compassion for the poor souls who went through the above process over and over again to find the derivatives of all the major functions, then compiled them into a master list for subsequent generations. It’s so much easier these days, because we know the derivative of t2 is 2t, for example, and even if we don’t, we can look it up.
Let’s revisit our problem again using this simpler process. We know we have a starting function of h(t)=−16t2+200. We want to know our instantaneous velocity when t = 1 second. So we take a derivative to find the relevant velocity function: .
Because we know the value of h, we can use substitution to rephrase the question as
Next we break up those parenthetical expressions to get
Now we move the −16 from the parentheses to get Where did that 0 come from? We took the derivative of 200, which is a constant, and one of the hard-and-fast rules in calculus is that the derivative of any constant is 0, because the derivative describes the rate of change and a constant doesn’t change.
All we have left to do is find the derivative of t2. This is where we can skip all those in-between steps, because we know that the derivative of t2 is 2t. That means we can plug 2t into The equation, so we end up with = - 16(2t).
We know that is equivalent to v(t). Multiply it out, and you end up with the same answer we got via our earlier belabored process: v(t) = −32t. Translated back into plain English: Our instantaneous velocity at our chosen point on the curve is −32 feet per second and because the velocity is equivalent to the slope of our tangent line, that slope is also −32t. (Our velocity is changing over time, so t must be included.)
Taking an Integral. The integral is the reverse of the derivative, so now we will reverse the question. This time we know our velocity as a function of