Q.A coat of paint of thickness 0.2 cm is applied to the faces of a cube whose edge is 10 cm. Use the differentials to find approximately how many cubic centimeters of paint is used to paint this cube. Also calculate the exact amount of paint used to paint this cube.
Concept understanding — Linear Approximation and Differentials
A nonlinear function is generally hard to evaluate exactly, but near any one point its graph looks almost like a straight line -- the tangent line at that point. This is the idea behind linear approximation.
Definition (Linear Approximation). Let f:(a,b)→R be differentiable and x0∈(a,b). The linear approximation L of f at x0 is
L(x)=f(x0)+f′(x0)(x−x0),∀x∈(a,b).
This is exactly the equation of the tangent line to y=f(x) at (x0,f(x0)). Because f is differentiable, f(x0+Δx)≈L(x0+Δx)=f(x0)+f′(x0)Δx when Δx is small, and the errorf(x)−L(x) shrinks to 0 faster than x→x0 (it is o(x−x0), a consequence of the definition of the derivative as a limit).
Tip
To linearly approximate an "ugly" value like 327.2 or (123)2/3: pick a nearby point x0 where the function and its derivative are easy to evaluate exactly (a perfect cube, a perfect fourth power, a multiple of 10, ...), compute f(x0) and f′(x0), and plug into L(x0+Δx)=f(x0)+f′(x0)Δx.
The differential. Writing Δx=dx, the tangent-line increment is called the differential of f:
df=f′(x)dx,equivalently df=f′(x)Δx.
Geometrically, Δf=f(x+dx)−f(x) is the actual rise along the curve, while df=f′(x)dx is the rise along the tangent line; for small dx, Δf≈df, but the two are generally not equal (only exactly equal when f is itself linear, f(x)=mx+c). df is a function of two independent quantities, x and dx -- not of x alone the way the derivative is.
Differentials of standard functions follow directly from the derivative rules, e.g. d(xn)=nxn−1dx, d(sinx)=cosxdx, d(ex)=exdx, d(logx)=x1dx. The algebraic properties of differentials mirror differentiation exactly:
Painting a 0.2 cm coat onto every face of the cube extends each pair of opposite faces outward by 0.2 cm each, so every edge effectively grows by 2(0.2)=0.4 cm; use the differential of V=x3 for the approximate paint volume, then compute the exact difference for comparison.
Step 1. Set up the cube's volume.V(x)=x3, base edge x0=10 cm.
Step 2. Determine the effective edge increment. A coat of thickness 0.2 cm on the OUTSIDE of each face means, along any one dimension, BOTH bounding faces gain 0.2 cm, so the edge length increases by a total Δx=2(0.2)=0.4 cm (new edge =10.4 cm).
Step 3. Approximate paint volume via the differential.V′(x)=3x2, so
dV=V′(x0)Δx=3(10)2(0.4)=3(100)(0.4)=120 cm3.
Step 4. Exact paint volume. Exact volume of paint = volume of the painted (larger) cube − volume of the original cube: …