009A Sample Final 1, Problem 8

Let

$y=x^{3}.$

(a) Find the differential $dy$ of $y=x^{3}$ at $x=2$.

(b) Use differentials to find an approximate value for $1.9^{3}$.

Foundations:
What is the differential $dy$ of $y=x^{2}$ at $x=1?$
Since  $x=1,$  the differential is  $dy=2xdx=2dx.$

Solution:

(a)

Step 1:
First, we find the differential $dy.$
Since $y=x^{3},$  we have
$dy\,=\,3x^{2}\,dx.$

Step 2:
Now, we plug $x=2$  into the differential from Step 1.
So, we get
$dy\,=\,3(2)^{2}\,dx\,=\,12\,dx.$

(b)

Step 1:
First, we find $dx$. We have  $dx=1.9-2=-0.1.$
Then, we plug this into the differential from part (a).
So, we have
$dy\,=\,12(-0.1)\,=\,-1.2.$

Step 2:
Now, we add the value for $dy$ to  $2^{3}$  to get an
approximate value of $1.9^{3}.$
Hence, we have
$1.9^{3}\,\approx \,2^{3}+-1.2\,=\,6.8.$

Final Answer:
(a)  $dy=12\,dx$
(b)  $6.8$

Return to Sample Exam