## Related Rates IB Maths

Discussions for the Core part of the syllabus. Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths HL Revision Notes

### Related Rates IB Maths

IB Mathematics HL– Calculus, Differentiation, Related Rates, Rate of change

How can we solve the following related rates word problem?
“A conical water tank with vertex down has a diameter of 6 meters at the top and a height of 10 meters. If water flows out of the tank at a rate of $1 m^3 /min$, how fast is the water level falling when the water is 2 meters deep?”

Thanks
Max

Posts: 0
Joined: Mon Jan 28, 2013 8:06 pm

### Re: Related Rates IB Maths

IB Maths HL – Calculus, Implicit Differentiation, Related rates, Rate of change.

The steps for solving Relates Rates word-problems are:

1. Carefully read the problem (what is asked for, and what is given)
2. Draw a diagram
3. Use appropriate variables to represent the quantities involved in the problem. Write down all equations that relate these variables and other given constants (e.g. Pythagorean Theorem, similar triangles, area, volume, proportions, trigonometric functions)
4. Differentiate both sides of the final equation with respect to time. This usually involves implicit differentiation.
5. Plug in given values and solve!!
Concerning you question,

Let $V$ represent the volume and $r$ represent the radius of the tank.

We know that $\frac{dV}{dt}=-1 m^3 /min$

The volume of the cone is given

$V=\frac{1}{3}\pi r^2 h$.

By similar triangles,
$\frac{3}{10}=\frac{r}{h}\Rightarrow r=\frac{3}{10} h$

So the volume of the cone can be written

$V=\frac{1}{3}\pi (\frac{3}{10} h)^2 h =$

$=\frac{9 \pi}{300} h^3$

Differentiating with respect to time:

$\frac{dV}{dt}= \frac{d}{dt}(\frac{9 \pi}{300} h^3) \Rightarrow$

$\frac{dV}{dt}= 3 \frac{9 \pi}{300} h^2 \frac{dh}{dt} \Rightarrow$

$\frac{dV}{dt}= \frac{9 \pi}{100} h^2 \frac{dh}{dt}$ (1)

Now plug everything in to equation (1) and solve for $\frac{dh}{dt}$ :

$-1= \frac{9 \pi}{100} (2)^2 \frac{dh}{dt} \Rightarrow$

$-1= \frac{9 \pi}{25} \frac{dh}{dt} \Rightarrow$

$\Rightarrow \frac{dh}{dt}= \frac{-25}{9 \pi}$ meters per minute

Hope these help!!
miranda

Posts: 268
Joined: Mon Jan 28, 2013 8:03 pm

### Re: Related Rates IB Maths

thanks miranda!
Max

Posts: 0
Joined: Mon Jan 28, 2013 8:06 pm