Differentiation Question 223
Question: If $ x=\frac{1-t^{2}}{1+t^{2}} $ and $ y=\frac{2at}{1+t^{2}} $ , then $ \frac{dy}{dx}= $
Options:
A) $ \frac{a(1-t^{2})}{2t} $
B) $ \frac{a(t^{2}-1)}{2t} $
C) $ \frac{a(t^{2}+1)}{2t} $
D) $ \frac{a(t^{2}-1)}{t} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ x=\frac{1-t^{2}}{1+t^{2}} $ and $ y=\frac{2at}{1+t^{2}} $
Differentiating with respect to t, we get $ \frac{dx}{dt}=\frac{(1+t^{2})(0-2t)-(1-t^{2})(0+2t)}{{{(1+t^{2})}^{2}}}=-\frac{4t}{{{(1+t^{2})}^{2}}} $
and $ \frac{dy}{dt}=\frac{(1+t^{2})2a-2at(2t)}{{{(1+t^{2})}^{2}}}=\frac{2a(1-t^{2})}{{{(1+t^{2})}^{2}}} $
Therefore $ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a(1-t^{2})}{-2t}; $
$ \therefore \frac{dy}{dx}=\frac{a(t^{2}-1)}{2t} $ .
BETA
