A) \[\frac{1}{2}\]
B) 0
C) -2
D) \[\infty \]
Correct Answer: B
Solution :
Here we have\[x=3{{t}^{2}}+1\] ... (i) \[\therefore \] \[\frac{dx}{dt}=6t\] \[y={{t}^{3}}-1\] ?. (ii) \[\frac{dy}{dt}=3{{t}^{2}}\] \[\therefore \] \[\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=3{{t}^{2}}\times \frac{1}{64}\] \[=\frac{t}{2}\] But from(i) when \[x=1\] we have \[1=3{{t}^{2}}+1\] \[3{{t}^{2}}=0\] \[\Rightarrow \] \[t=0\] \[\therefore \] When \[x=1\] then \[t=0\] \[\therefore \] \[\frac{dy}{dx}=0\]You need to login to perform this action.
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