The derivative of a function gives us the slope of a tangent line to that function. We can use this tangent line to estimate how much the output (\(y\) values) will change if we change the input (\(x\)-value). If we rewrite the notation \(\ds\frac{dy}{dx}=f'\) in the form \(dy=f' dx\text{,}\) then we can read this as “A small change in \(y\) (called \(dy\)) equals the derivative (\(f'\)) times a small change in \(x\) (called \(dx\)).”
We call \(dx\) the differential of \(x\text{.}\) If \(f\) is a function of \(x\text{,}\) then the differential of \(f\) is \(df = f'(x) dx\text{.}\) Since we often write \(y=f(x)\text{,}\) we'll interchangeably use \(dy\) and \(df\) to represent the differential of \(f\text{.}\)
We will often refer to the differential notation \(dy=f'dx\) as “a change in the output \(y\) equals the derivative times a change in the input \(x\text{.}\)”
If \(f(x) = x^2\ln(3x+2)\) and \(g(t) = e^{2t}\tan(t^2)\) then compute \(df\) and \(dg\text{.}\)
Most of higher dimensional calculus can quickly be developed from differential notation. Once we have the language of vectors and matrices at our command, we will develop calculus in higher dimensions by writing \(d\vec y = Df(\vec x) d\vec x\text{.}\) Variables will become vectors, and the derivative will become a matrix.
This problem will help you see how the notion of differentials is used to develop equations of tangent lines. We'll use this same idea to develop tangent planes to surfaces in 3D and more.
Consider the function \(y=f(x) = x^2\text{.}\) This problem has multiple steps, but each is fairly short.
The linearization of a function is just an equation of the tangent line where you solve for \(y\text{.}\)
Find the differential of \(y\) with respect to \(x\text{.}\)
Draw a graph of \(f(x)\text{.}\) Place a dot at the point \((3,9)\) and label it on your graph. Sketch a tangent line to the graph at the point \((3,9)\) on the same axes. Place another dot on the tangent line up and to the right of (3,9). Label the point \((x,y)\text{,}\) as it will represent any point on the tangent line.
Using the two points \((3,9)\) and \((x,y)\text{,}\) compute the slope of the line connecting these two points. Your answer should involve \(x\) and \(y\text{.}\)
What is the rise (i.e, the change in \(y\) called \(dy\))? What is the run (i.e, the change in \(x\) called \(dx\))?
We know the slope of the tangent line is the derivative \(f'(3)=6\text{.}\) We also know the slope from the previous part. These two must be equal. Use this fact to give an equation of the tangent line to \(f(x)\) at \(x=3\text{.}\) (Hint: do NOT simplify to slope-intercept form)
How is the equation for the tangent line related to the differentials \(dy\) and \(dx\text{?}\)
The manufacturer of a spherical storage tank needs to create a tank with a radius of 3 m. Recall that the volume of a sphere is \(V(r) = \frac{4}{3}\pi r^3\text{.}\) No manufacturing process is perfect, so the resulting sphere will have a radius of 3 m, plus or minus some small amount \(dr\text{.}\) The actual radius will be \(3+dr\text{.}\)
Find the differential \(dV\text{.}\) This should be a formula with \(dV\text{,}\) \(r\text{,}\) and \(dr\text{.}\)
If the actual radius is 3.02...
What is \(r\text{?}\)
What is \(dr\text{?}\)
What is \(dV\) equal to? What does \(dV\) tell you about the volume of the manufactured sphere?
A forest ranger needs to estimate the height of a tree. The ranger stands 50 feet from the base of tree and measures the angle of elevation to the top of the tree to be about 60\(^\circ\text{.}\) If this angle of 60\(^\circ\) is correct, then what is the height of the tree? If the ranger's angle measurement could be off by as much as \(5^\circ\text{,}\) then how much could his estimate of the height be off? Use differentials to give an answer.