Thread: Some questions from old exams??

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

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

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

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/$\displaystyle ft^2$. 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 $\displaystyle z=e^{-(x^2+y^2)}$ 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. =)

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: .$\displaystyle V \;=\;\tfrac{\pi}{3}r^2h$

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

The percentage error is: .$\displaystyle 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: .$\displaystyle \begin{Bmatrix}r \:=\: 40 & dr \:\leq\: 0.8 \\ h \:=\:20 & dh \:\leq\:0.4 \end{Bmatrix}$


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

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

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

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, $\displaystyle \frac{xy^2}{x^4y^4}=\frac{1}{x^3 y^2}$ ..

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

Problem 4 : Shout be straightforward for you!

Originally Posted by General

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

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

thx to our lecturer, dont know what will happen.

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 ?!

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

Originally Posted by Lafexlos

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

Yes ..

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