In S. Thompson's Calculus Made Easy (pg. 31) He talks about an example which he writes, and I quote:

(4) The volume of a cylinder of radius $$r$$ and height $$h$$ is given by the formula $$V = πr²h$$. Find the rate of variation of volume with the radius when $$r = 5.5 in.$$ and $$h = 20 in.$$ If $$r = h$$, find the dimensions of the cylinder so that a change of 1 in. in radius causes a change of 400 cub. in. in the volume.

The rate of variation of $$V$$ with regard to $$r$$ is $$dV/dr = 2πrh$$ If $$r = 5.5 in.$$ and $$h = 20 in.$$ this becomes $$690.8$$. It means that a change of radius of 1 inch will cause a change of volume of $$690.8$$ cub. inch. This can easily be verified, for the volumes with $$r = 5$$ and $$r = 6$$ are $$1570 cub. in.$$ and $$2260.8 cub. in.$$ respectively, and $$2260.8 – 1570 = 690.8$$ Also, if $$r = h$$, $$dV/dr = 2πr² = 400$$ and $$r = h = √(400/2π) = 7.98 in$$

Forget about the second part for now. Considering only the first part, Does that mean for every consecutive increase in radius, the increase in volume is $$690.8 in³$$? Because if I use radii other than the ones used in the example, the increase in volume is different every time. For instance if radius is $$30$$ in., the volume is $$56540 in³$$ and radius $$31$$ gives volume $$60530.8 in³$$ the difference being $$3990.8 in³$$ which contradicts the statement

It means that a change of radius of 1 inch will cause a change of volume of $$690.8$$ cub. inch.

Is there anything I'm missing? Or am I interpreting it the wrong way? Please explain?

Throw. This. Book. Away.

The author shows very little fundamental understanding of calculus. Let's try the same arguments with the volume of a sphere: $$V=\frac43\pi r^3\ ,\qquad \frac{dV}{dr}=4\pi r^2\ .$$ If $$r=5.5$$ and we find the difference in volumes for $$r=5$$ and $$r=6$$, exactly as in the book, we get $$\frac{dV}{dr}=121\pi\ ,\qquad V(6)-V(5)=(121\tfrac13)\pi\ ,$$ which are clearly not the same, though as we would expect from a proper understanding of calculus, they are approximately the same.

So, why did it work for the cylinder? Purely because the volume formula for the cylinder involves an $$r^2$$. We can easily do the algebra for any value of $$r$$, varying it by $$\frac12$$ each way as in the book: $$V(r+\tfrac12)-V(r-\tfrac12)=\pi\bigl((r+\tfrac12)^2-(r-\tfrac12)^2\bigr)h =2\pi rh=\frac{dV}{dr}\ .$$ This is purely algebra, nothing to do with calculus. And more importantly, as shown in the sphere example, it won't work for other examples.

A small change from $$r$$ to $$r+\delta$$ will cause a change in $$V$$ of $$approximately$$ $$2\pi rh\delta .$$

A small change from $$r^*=r+\delta$$ to $$r^*-\delta =r$$ will cause a change in $$V$$ of $$approximately$$ $$-2\pi r'h\delta.$$

But (if $$r\ne 0\ne h$$) $$2\pi rh\delta \ne 2\pi r'h\delta$$ unless $$\delta =0.$$

And note that $$2\pi rh \delta$$ is different for different values of $$r$$.

The book is misleading because it refers to these formulas for changes as if they were exact and not approximate.

You are absolutely right in thinking.

The change itself depends on the radius. This is not surprising. Only in case of linear functions like ax+b, the change doesn't depend on the value of parameter. But, in general, the change depends on the parameters/variables. That is why it is specifically mentioned in the question $$r=h$$. If every function was like that(linear), life would have been so easy! And, that is where the differentiation and integration are helpful. If things were constant, then we would not even need calculus!

• So what does that mean? Is there anything in the question that remains constant? Why does he say "It means that a change of radius of 1 inch will cause a change of volume of $690.8$ cub. inch."? If that's wrong, is there something that remains constant throughout? – user231094 May 16 at 1:56
• The height of cylinder is constant for this question – Tojrah May 16 at 2:17