MCQ
If $\text{x}=\text{at}^2,\text{y}=2\text{at}$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}=$
- A$-\frac{1}{\text{t}^2}$
- B$\frac{1}{2\text{at}^3}$
- C$-\frac{1}{\text{t}^3}$
- ✓$-\frac{1}{2\text{at}^3}$
✓
Answer
Correct option: D.
$-\frac{1}{2\text{at}^3}$
$\text{x}=\text{at}^2,\text{y}=2\text{at}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\text{a}}{2\text{at}}$
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{d}}{\text{dt}}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\frac{\text{dx}}{\text{dt}}}$
$=\frac{\frac{-1}{\text{t}^2}}{2\text{at}}=\frac{-1}{2\text{at}^3}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\text{a}}{2\text{at}}$
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{d}}{\text{dt}}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\frac{\text{dx}}{\text{dt}}}$
$=\frac{\frac{-1}{\text{t}^2}}{2\text{at}}=\frac{-1}{2\text{at}^3}$
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