logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

Choose the correct answer in the slope of the tangent to the curve \(x = t^2 + 3t – 8, y = 2t^2 – 2t – 5\) at the point \((2,– 1)\) is

\[(A)\; \frac{22}{7} \quad (B)\; \frac{6}{7} \quad (C) \;\frac{7}{6} \quad (D) \;\frac{-6}{7}\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
The curve is $x=t^2+3t-8,y=2t^2-2t-5$------(1)
Put $x=2$
$2=t^2+3t-8$
$t^2+3t-8-2=0$
$t^2+3t-10=0$
$t^2+5t-2t-10=0$
$t(t+5)-2(t+5)=0$
$(t+5)(t-2)=0$
$t=-5,2$
Step 2:
Put $t=2$ in
$y=2t^2-2t-5$
$\;\;=2(2)^2-2(2)-5$
$\;\;=8-4-5$
$\;\;=-1$
At $x=2,y=-1,t=2$
Step 3:
Differentiating (1) with respect to $t$
$\large\frac{dx}{dt}$$=2t+3$
$\large\frac{dy}{dt}$$=4t-2$
$\large\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$
$\large\frac{dy}{dx}=$$4t-2\times \large\frac{1}{2t+3}$
$\quad\;\;=\large\frac{4t-2}{2t+3}$
Step 4:
At $t=2$
$\large\frac{dy}{dx}=\frac{4\times 2-2}{2\times 2+3}$
$\qquad=\large\frac{6}{7}$
Part (B) is the correct answer.
answered Aug 13, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...