Applications-Of-Derivatives Question 459
Question: The slope of the tangent to the curve $ x=3t^{2}+1,y=t^{3}-1 $ at $ x=1 $ is
[Karnataka CET 2003]
Options:
A) 0
B) $ \frac{1}{2} $
C) $ \infty $
D) $ -2 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ x=3t^{2}+1,,y=t^{3}-1 $ \ $ \frac{dx}{dt}=6t, $ $ \frac{dy}{dt}=3t^{2} $ Now $ \frac{dy}{dx}=( \frac{\frac{dy}{dt}}{\frac{dx}{dt}} ) $ = $ \frac{3t^{2}}{6t}=\frac{t}{2} $ For $ x=1 $ , $ 3t^{2}+1=1\Rightarrow t=0 $
Þ Slope = $ \frac{0}{2}=0 $ .
BETA
