Answer:
The value of [tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] at t = √2 is [tex]\frac{1}{2 \times \sqrt{2}}[/tex] .
Step-by-step explanation:
Given as :
x = t³ - 3 t
y = 3 t²
We have to calculate [tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] at t = √2
Now,
[tex]\frac{\mathrm{d} x}{\mathrm{d} t}[/tex] = [tex]\frac{\mathrm{d} (t³ - 3 t)}{\mathrm{d} t}[/tex]
Or, [tex]\frac{\mathrm{d} x}{\mathrm{d} t}[/tex] = [tex]\frac{\mathrm{d} t³}{\mathrm{d} t}[/tex] - 3 [tex]\frac{\mathrm{d} t}{\mathrm{d} t}[/tex]
Or, [tex]\frac{\mathrm{d} x}{\mathrm{d} t}[/tex] = 3 t² - 3 ......A
Again
[tex]\frac{\mathrm{d} y}{\mathrm{d} t}[/tex] = [tex]\frac{\mathrm{d} (3 t²)}{\mathrm{d} t}[/tex]
Or, [tex]\frac{\mathrm{d} y}{\mathrm{d} t}[/tex] = 3 × 2 t
Or, [tex]\frac{\mathrm{d} y}{\mathrm{d} t}[/tex] = 6 t ......B
So,
[tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] = [tex]\frac{\frac{\partial y}{\partial t}}{\frac{\partial x}{\partial t}}[/tex]
Or, [tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] = [tex]\dfrac{3 t^{2} - 3 }{6 t}[/tex]
Or, [tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] = [tex]\frac{3(t^{2}-1)}{6 t}[/tex]
Or, [tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] = [tex]\frac{(t^{2}-1)}{2 t}[/tex]
Or, [tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] at t = √2
i.e [tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] = [tex]\frac{(\sqrt{2}^{2}-1)}{2 \times \sqrt{2}}[/tex]
Or, [tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] = [tex]\frac{(2-1)}{2 \times \sqrt{2}}[/tex]
Or, [tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] = [tex]\frac{1}{2 \times \sqrt{2}}[/tex]
Hence, The value of [tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex] at t = √2 is [tex]\frac{1}{2 \times \sqrt{2}}[/tex] . Answer
