Differentiating cube diagonal growth by time

Key concept: This problem is about converting a geometric formula into a time-derivative formula. The key point is that the cube’s space diagonal is proportional to its side length, so the rate follows directly after differentiation.

Step by step

Identify the geometric formula

Let $x$ be the side length of the cube and $z$ be the space diagonal. The diagonal of a cube is

$z=\sqrt{3}\,x.$

This is not an equation you need to derive from scratch every time; it comes from the 3D version of the Pythagorean theorem. In a cube, the space diagonal uses three equal edges, so its length is the square root of $1^2+1^2+1^2=3$ times the side.

Differentiate the equation with respect to time

Because both $x$ and $z$ may change over time, treat them as functions of $t$:

$\frac{dz}{dt}=\frac{d}{dt}(\sqrt{3}\,x).$

Since $\sqrt{3}$ is just a constant, it stays outside the derivative:

$\frac{dz}{dt}=\sqrt{3}\,\frac{dx}{dt}.$

That is the marked step. There is no extra product rule here because $\sqrt{3}$ is constant, not a variable.

Substitute the given rate

If the side length increases at $5$ cm/hr, then

$\frac{dx}{dt}=5.$

So the diagonal’s rate is

$\frac{dz}{dt}=\sqrt{3}(5)=5\sqrt{3}\text{ cm/hr}.$

Why the step looks simpler than the rectangle case

In the rectangle problem, you had two changing variables tied together by a square-law equation, so differentiating produced multiple terms. Here, the diagonal is already written as a constant multiple of the side. Differentiation is therefore immediate: a constant multiple stays a constant multiple. No chain rule complication appears beyond recognizing that $x$ depends on time.

A quick check

The diagonal should grow faster than the side because it stretches across the cube. Since $\sqrt{3}\approx1.732$, the diagonal rate should be larger than $5$ but not dramatically larger. The answer $5\sqrt{3}$ matches that expectation.

Pitfall alert

Students often overcomplicate this by trying to re-derive the cube diagonal every time or by applying the product rule where it is not needed. The only variable is $x$, while $\sqrt{3}$ is constant. Another mistake is mixing up face diagonal and space diagonal: the face diagonal is $x\sqrt{2}$, but the space diagonal is $x\sqrt{3}$. Using the wrong diagonal formula changes the rate immediately.

Try different conditions

If the cube side were increasing at $r$ cm/hr instead of 5 cm/hr, the same derivation gives $\frac{dz}{dt}=\sqrt{3}\,r$. If the problem asked for the rate of the face diagonal rather than the space diagonal, you would use $z=x\sqrt{2}$ and get $\frac{dz}{dt}=\sqrt{2}\,\frac{dx}{dt}$. The structure is the same; only the geometric constant changes.

Further reading

space diagonal, constant multiple rule, related rates