Some questions from old exams??

**Lafexlos** · Mar 25th 2010, 06:13 AM

1)
a) $\lim_{(x,y)\rightarrow(0,0)}\frac{sin(xy)}{x}$

b) $\lim_{(x,y)\rightarrow(0,0)}\frac{xy^2}{x^4y^4}$

2) Without using distance formula, find the distance from the point (0,1,1) to the plane $4y+3z=-12$

3) A closed box to be found to have length 2ft, witdh 4ft and hight 3ft, where the measurement of each dimension ismade with a maximum possible error of +-0.02ft. The top of the box is made from material that costs only 1.50 Dollars/ $ft^2$ . What is the maximum error involved in the computation of the cost of the box?

4) Find an equation for the tangent plane to the surface $z=e^{-(x^2+y^2)}$ at (0,0,1).

5) A right circular cone is measured and is found to have base radius r=40cm and altitude h=20cm. If it is known that each measurement is accurate to within %2, what is the maximum percentage error in the measurement of the volume?

These are five of the problems which i couldn't solve. i'll try others based on answers of these.
Any help is appriciated. =)

**Soroban** · Mar 25th 2010, 07:59 AM

Hello, Lafexlos!

Here's #5 . . .

5) A right circular cone is found have base radius r = 40 cm and height h = 20 cm.
If it is known that each measurement is accurate to within 2%,
what is the maximum percentage error in the measurement of the volume?

The volume of a cone is: . $V \;=\;\tfrac{\pi}{3}r^2h$

Hence: . $dV \;=\;\tfrac{\pi}{3}r^2dh + \tfrac{2\pi}{3}rh\,dr \;=\;\tfrac{\pi}{3}r(r\,dh + 2h\,dr)$

The percentage error is: . $E \;=\;\frac{dV}{V} \;=\;\frac{\frac{\pi}{3}r(r\,dh + 2h\,dr)}{\frac{\pi}{3}r^2h} \;=\; \frac{r\,dh + 2h\,dr}{rh}$ .[1]

We are given: . $\begin{Bmatrix}r \:=\: 40 & dr \:\leq\: 0.8 \\ h \:=\:20 & dh \:\leq\:0.4 \end{Bmatrix}$

Substitute into [1]: . $E \;\leq\;\frac{(40)(0.4) + 2(20)(0.6)}{(40(20)} \;=\;0.06 \;=\;\boxed{6\%}$

**Lafexlos** · Mar 25th 2010, 12:14 PM

Thanks Soroban. =)
Still need help at* others.

*[on, or whatever. still dont know prepositions. :A]

**General** · Mar 25th 2010, 12:23 PM

For the first limit, when y=0 : the limit = 0 .. when y=2 .. the limit = 2
so the limit does not exist.

for the second limit, $\frac{xy^2}{x^4y^4}=\frac{1}{x^3 y^2}$ ..

So what will happen when $(x,y) \rightarrow (0,0)$ ?

Problem 4 : Shout be straightforward for you!

**Lafexlos** · Mar 25th 2010, 12:41 PM

Originally Posted by General

for the second limit, $\frac{xy^2}{x^4y^4}=\frac{1}{x^3 y^2}$ ..

So what will happen when $(x,y) \rightarrow (0,0)$ ?

thx to our lecturer, dont know what will happen.

**General** · Mar 25th 2010, 12:41 PM

Originally Posted by Lafexlos

thx to our lecturer, dont know what will happen.

When you substitute x=0 and y=0 .. what will happen to the fraction ?!

**Lafexlos** · Mar 25th 2010, 12:44 PM

infinite.
Hence it doesn't exist too because of limit goes infinite?

**General** · Mar 25th 2010, 12:45 PM

Originally Posted by Lafexlos

infinite.
Hence it doesn't exist too because of limit goes infinite?

Yes ..

**Lafexlos** · Mar 25th 2010, 12:49 PM

ok. thanks.
#1 & #5 solved. Try to solve #4 again. So #2 && #3 remain.

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