Answer to Question #131053 in Calculus for Break Down

Question

A right circular cone is held with its apex facing downwards and is inscribed in a sphere with fixed radius 𝑘 cm within the interval 1 cm ≤ 𝑘 ≤ 5 cm and the distance from the base of the cone to the center of the sphere is 𝑥 cm.

a. Evaluate the maximum volume of the cone by assigning 𝑘 to a value within the given interval. Discuss the difference of use between the first derivative test and second derivative test to optimize the volume.

b. Water is poured into the cone at a rate of 10 m3s−1. Find the rate at which the water level is rising when the depth of the water is (𝑥 + 𝑘) − 3.

Answer 1

a.

$$V={1\over3}\pi r^2 h$$

Use the Pythagorean Theorem to the right triangle $\Delta BOD$

$$k^2=x^2+r^2 , 0\leq x\leq k$$

Then

$$r^2=k^2-x^2$$

The height is

$$h=k+x \ \ or\ \ h=k-x$$

Let $|u|=x, -k\leq u\leq k.$ Then

$$r^2=k^2-u^2, h=k+u$$

$$V(u)={1\over3}\pi (k^2-u^2)(k+u)$$

$$V(u)={1\over3}\pi(k^3+k^2u-ku^2-u^3)$$

Find the first derivative with respect to $u$

$$V'(u)={1\over3}\pi(0+k^2-2ku-3u^2)=$$

$$={1\over3}\pi(k^2-2ku-3u^2)$$

Find the critical number(s)

$$V'(u)=0=>{1\over3}\pi(k^2-2ku-3u^2)=0$$

$$-3u^2-2ku+k^2=0$$

$$u=\dfrac{2k\pm \sqrt{(-2k)^2-4(-3)(k^2)}}{2(-3)}=\dfrac{-1\pm 2}{3}\cdot k$$

$$u_1=-k, u_2={1\over 3}k$$

Critical numbers: $-k, \dfrac{1}{3}k.$

First Derivative Test

If $<u<-k, V'(u)<0, V(u)$ decreases.

If $-k<u<\dfrac{1}{3}k, V'(u)>0, V(u)$ increases.

If $u>\dfrac{1}{3}k, V'(u)<0, V(u)$ decreases.

$-k\leq u\leq k$

$$V(-k)={1\over3}\pi(k^3+k^2(-k)-k(-k)^2-(-k)^3)=0$$

$$V(\dfrac{1}{3}k)={1\over3}\pi(k^3+k^2(\dfrac{1}{3}k)-k(\dfrac{1}{3}k)^2-(\dfrac{1}{3}k)^3)=$$

$$=\dfrac{32}{81}\pi k^3(cm^3)$$

$$V(k)={1\over3}\pi(k^3+k^2(k)-k(k)^2-(k)^3)=0$$

The volume has the maximum with value of $\dfrac{32}{81}k^3$ cm³ for $-k\leq u\leq k$ at $u=\dfrac{1}{3}k$ .

Find the second derivative with respect to $u$

$$V''(u)={1\over3}\pi(0-2k-6u)=-{2\over3}\pi(k+3u)$$

Second Derivative Test

Critical numbers: $-k, \dfrac{1}{3}k.$

$$V''(-k)=-{2\over3}\pi(k+3(-k))={4\over3}\pi k>0, k>0$$

$$V''(\dfrac{1}{3}k)=-{2\over3}\pi(k+3(\dfrac{1}{3}k))=-{4\over3}\pi k<0, k>0$$

The volume has the maximum with value of $\dfrac{32}{81}k^3$ cm³ for $-k\leq u\leq k$ at $u=\dfrac{1}{3}k$ .

Therefore $V_{max}=\dfrac{32}{81}k^3$ cm³ when we have case 1(Figure 1).We see than $V_{max}$ increases when $k$ increases.

$$V_{max}=\dfrac{4000}{81}cm^3, k=5 \ cm, x=\dfrac{5}{3}\ cm$$

The difference is that the firstrivative test always determines whether a function has a local maximum, a local minimum, or neither; however, the second derivative test fails to yield a conclusion when y'' is zero at a critical value. If that is the case, you will have to apply the first derivative test to draw a conclusion.

b.

Let $q(t)=$ the height of the cone formed by water at time $t$ , $p(t)=$ the radius of the base of the cone formed by water at time $t$

We have that the volume of water is

$$V(t)={1\over3}\pi p^2q$$

From the similar triangles

$$\dfrac{r}{h}=\dfrac{p}{q}$$

Solve for $p$

$$p=\dfrac{r}{h}\cdot q$$

Substitute

$$V(t)={1\over3}\pi (\dfrac{r}{h}\cdot q)^2q ={\pi r^2\over3h^2}q^3$$

$$V(t)={\pi (k^2-u^2)\over3(k+u)^2}q^3$$

$$V(t)={\pi (k-u)\over3(k+u)}q^3$$

Differentiate both sides with respect to $t$

$$\dfrac{d}{dt}\big(V\big)=\dfrac{d}{dt}\big({\pi (k-u)\over3(k+u)}q^3\big)$$

Use the Chain Rule

$$\dfrac{dV}{dt}={\pi (k-u)\over k+u}q^2\dfrac{dq}{dt}$$

Solve for $\dfrac{dq}{dt}$

$$\dfrac{dq}{dt}=\dfrac{k+u}{\pi(k-u)q^2}\cdot \dfrac{dV}{dt}$$

Given

$\dfrac{dV}{dt}=10m^3\cdot s^{-1}=10^7 cm^3\cdot s^{-1}$

$q=x+k-3>0$

I don't believe that $\dfrac{dV}{dt}=10m^3\cdot s^{-1}.$ This is the very big rate.

Suppose $\dfrac{dV}{dt}=10cm^3\cdot s^{-1}$

Then for $q=x+k-3>0$

In case 1 (FIgure 1)

$u=x$

$$\dfrac{dq}{dt}=\dfrac{k+x}{\pi(k-x)(x+k-3)^2}\cdot (10\ cm\cdot s^{-1})$$

In case 2 (FIgure 2)

$u=-x$

$$\dfrac{dq}{dt}=\dfrac{k-x}{\pi(k+x)(x+k-3)^2}\cdot (10\ cm\cdot s^{-1})$$