Consider the curve \( y=\frac{3}{2} x^{3}+4 x \) Show that the equatio

1.	Consider the curve \( y=\frac{3}{2} x^{3}+4 x \) Show that the equation of the tangent to the Curve at \( x=2 \) is given by \( y=22 x-24 \) ii Find the equation of the normal to the curve at \( x=2 \)
Answer» Given curve is : y = \(\frac{3}{2}x^3+4x\) ....(1) ∴ Slope of tangent, \(\frac{dy}{dx}\) = \(\frac{9}{2}x^2+4\) ∴ Slope of tangent to given curve at x = 2 is : \((\frac{dy}{dx})\)_{(x = 2)} = \(\frac{9}{2}\)x 4 + 4 = 18 + 4 = 22. Put x = 2 in equation (1), we get y = \(\frac{3}{2}\) x 8 + 8 = 12 + 8 = 20. Hence, (2,20) is point on the given curve when x = 2. Now, Equation of tangent to curve at x = 2 is : y - 20 = \((\frac{dy}{dx})\)_{(x = 2)} (x-2) ⇒ y - 20 = 22(x-2) ⇒ y - 20 = 22(x-2) ⇒ y = 22x - 44 + 20 ⇒ y = 22x - 24. Slope of normal to given curve at x = 2 is : \(\frac{-1}{(\frac{dy}{dx})_{(x=2)}}\) = \(\frac{-1}{22}\). ∴ Equation of normal to curve at (2,20) is : y - 20 = ( slope of normal)(x - 2) ⇒ y - 20 = \(\frac{-1}{22}(x-2)\) ⇒ 22y - 440 = - (x - 2) ⇒ x - 2 + 22y - 440 = 0 ⇒ x + 22y - 442 = 0 Hence, Equation of normal to the curve at x = 2 is x + 22y- 442 = 0.

Consider the curve \( y=\frac{3}{2} x^{3}+4 x \) Show that the equation of the tangent to the Curve at \( x=2 \) is given by \( y=22 x-24 \) ii Find the equation of the normal to the curve at \( x=2 \)

Answer»

Given curve is :

y = \(\frac{3}{2}x^3+4x\) ....(1)

∴ Slope of tangent,

\(\frac{dy}{dx}\) = \(\frac{9}{2}x^2+4\)

∴ Slope of tangent to given curve at x = 2 is :

\((\frac{dy}{dx})\)_{(x = 2)}

= \(\frac{9}{2}\)x 4 + 4

= 18 + 4

= 22.

Put x = 2 in equation (1), we get

y = \(\frac{3}{2}\) x 8 + 8

= 12 + 8

= 20.

Hence,

(2,20) is point on the given curve when x = 2.

Now,

Equation of tangent to curve at x = 2 is :

y - 20 = \((\frac{dy}{dx})\)_{(x = 2)} (x-2)

⇒ y - 20 = 22(x-2)

⇒ y = 22x - 44 + 20

⇒ y = 22x - 24.

Slope of normal to given curve at x = 2 is :

\(\frac{-1}{(\frac{dy}{dx})_{(x=2)}}\) = \(\frac{-1}{22}\).

∴ Equation of normal to curve at (2,20) is :

y - 20 = ( slope of normal)(x - 2)

⇒ y - 20 = \(\frac{-1}{22}(x-2)\)

⇒ 22y - 440 = - (x - 2)

⇒ x - 2 + 22y - 440 = 0

⇒ x + 22y - 442 = 0

Hence,

Equation of normal to the curve at x = 2 is x + 22y- 442 = 0.