Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Answer
Comment
Share
Q)

Using differentials, find the approximate value of each of the following up to 3 places of decimal. $(ii)\;\sqrt{49.5}$

This is second part of multipart q1

1 Answer

Comment
A)
Toolbox:
  • Let $y=f(x)$
  • $\Delta x$ denote a small increment in $x$
  • $\Delta y=f(x+\Delta x)-f(x)$
  • $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
Step 1:
Let $\Delta y=\sqrt{x+\Delta x}-\sqrt x$
$\Delta y=\sqrt{x+\Delta x}-\sqrt x$
$\quad\;=\sqrt{49+.5}-\sqrt {49}$
$\quad\;=\sqrt{49.5}-\sqrt {49}$
$\quad\;=\sqrt{49.5}-7$
$\sqrt{49.5}=7+\Delta y$------(1)
Step 2:
$dy$ is the approximate value of $\Delta y$
$dy=\big(\large\frac{dy}{dx}\big)$$\Delta x$
We know that $y=\sqrt x$
$\large\frac{dy}{dx}=\frac{1}{2\sqrt x}$[Differentiating with respect to $x$]
$dy=\large\frac{dy}{dx}$$.\Delta x$
$\quad\;=\large\frac{1}{2\sqrt x}$$.\Delta x$
$\quad\;=\large\frac{1}{2\sqrt {49}}$$.0.5$
$\quad\;=\large\frac{1}{2\times 7}$$.0.5$
$\quad\;=\large\frac{1}{14}$$\times 0.5$
$\quad\;=0.036$
Step 3:
$\sqrt{49.5}=7+\Delta y$[From equation (1) we have]
Now substitute the value of $dy$ in (1)
$\sqrt{49.5}=7+0.036$
$\qquad\;\;=7.036$
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...