Application of Derivatives Test - 11

Let the side of an equilateral triangle be x cm.

\(\therefore\)  Area of equilateral triangle, A = \(\frac{\sqrt 3}{4} x^2\)  ....(i)

Also, \(\frac{dx}{dt}\) = 2cm/s

On differentiating Eq. (i) w.r.t. t, we get

\(\frac{dA}{dt} = \frac{\sqrt 3}{4}.2x.\frac{dx}{dt}\)

= \(\frac{\sqrt 3}{4}.2.10.2[\because x = 10 \) and \(\frac{dx}{dt} = 2]\)

= 10 \(\sqrt 3\) \(cm^2/s\)

We have, y = \(x^{\frac{1}{5}}\)

⇒ \(\frac{dy}{dx} = \frac{1}{5}x^{{\frac{1}{5}} -1}\) = \(\frac{1}{5}x^{\frac{-4}{5}}\)

\(\therefore (\frac{dy}{dx})_{(0,0)} = \frac{1}{5} \times (0)^{\frac{-4}{5}} \)

= \(\infty\)

So, the curve y = \(x^{\frac{1}{5}}\) has vertical tangent at (0, 0), which is parallel to Y-axis.

We have, the equation of the curve is \(3x^2 - y^2 = 8\) ....(i)

Also, the given equation of the line is x + 3y = 8

⇒ 3y = 8 - x

⇒ y = \(- \frac{x}{3} + \frac{8}{3}\)

Thus, slope of the line is \(-\frac{1}{3}\) which should be equal to slope of the equation of normal to the curve.

On differentiating Eq. (i) w.r.t. x, we get

\(6x -2y \frac{dy}{dx} = 0\)

⇒ \(\frac{dy}{dx} = \frac{6x}{2y} = \frac{3x}{y}\) = Slope of the curve

Now, slope of normal to the curve = \(-\frac{1}{(\frac{dy}{dx})}\)

= \(-\frac{1}{\frac{3x}{y}} = -\frac{y}{3x}\)

\(\therefore\) - \((\frac{y}{3x}) = -\frac{1}{3}\)

⇒ -3y = -3x

⇒ y = x

On substituting the value of the given equation of the curve, we get

\(3x^2 - x^2 = 8\)

⇒ \(x^2 = \frac{8}{2}\)

⇒ x = \(\pm 2\)

For x = 2, 3\((2)^2 - y^2 = 8\)

⇒ \(y^2 = 4\)

⇒ y = \(\pm 2\)

and for x = -2, 3\((2)^2 - y^2 = 8\)

⇒ y = \(\pm 2\)

So, the points at which normal are parallel to the given line are (\(\pm 2\), \(\pm 2\)).

Hence, the equation of normal at (\(\pm 2\), \(\pm 2\)) is

\(y - (\pm 2) = -\frac{1}{3}[x - (\pm 2)]\)

⇒ 3[y - (\(\pm 2\))] = -[x - (\(\pm 2\))]

⇒ x + 3y \(\pm\) 8 = 0

We have, \(ay + x^2 = 7\) and \(x^3 = y\)

On differentiating w.r.t. x. in both equations, we get

\(a.\frac{dy}{dx} + 2x = 0\) and \(3x^2 = \frac{dy}{dx}\)

⇒ \(\frac{dy}{dx} = -\frac{2x}{a} \) and \(\frac{dy}{dx} = 3x^2\)

⇒ \((\frac{dy}{dx})_{(1,1)} = \frac{-2}{a} = m_1\)

and \((\frac{dy}{dx})_{(1,1)}\) = 3.1 = 3 = \(m_2\)

Since, the curves cut orthogonally at (1, 1)

\(\therefore m_1.m_2 = -1\)

⇒ \((\frac{-2}{a}).3 = -1\)

\(\therefore\) a = 6

We have, \(y = x^4 - 10\)

⇒ \(\frac{dy}{dx} = 4x^3\)

and \(\Delta x \) = 2.00 - 199 = 0.01

\(\therefore \Delta y = \frac{dy}{dx} \times \Delta x\)

= \(4x^3 \times \Delta x\)

= 4 x \(2^3\) x 0.01

= 32 x 0.01 = 0.32

So, the approximate change in y is 0.32.

We have, equation of the curve \(y(1 + x^2) = 2 - x\) ....(i)

\(\therefore\) y.(0 + 2x) + (1 + \(x^2). \frac{dy}{dx}\) = 0 - 1 [on differentiating w.r.t.x]

⇒ 2xy + (1 + \(x^2\))\(\frac{dy}{dx} = -1\)

⇒ \(\frac{dy}{dx} = \frac{-1 - 2xy}{1 + x^2}\) ...(ii)

Since, the given curve passes through X-axis i.e., y = 0.

\(\therefore\) 0(1 + \(x^2\)) = 2 - x [using Eq.(i)]

⇒ x = 2

So, the curve passes through the point (2, 0).

\(\therefore\) \((\frac{dy}{dx})_{(2,0)}\) = \(\frac{-1 - 2 \times 0}{1 + 2^2}\)

= \(-\frac{1}{5}\) = Slope of the curve

\(\therefore\) Slope of tangent to the curve = \(-\frac{1}{5}\)

\(\therefore\) Equation of tangent of the curve passing through (2, 0) is

y - 0 = \(-\frac{1}{5}\)(x - 2)

⇒ 5y = -x + 2

⇒ 5y + x = 2

The given equation of curve is

\(y = x^3 - 12x + 18\)

\(\therefore \frac{dy}{dx} = 3x^2 - 12\) [on differentiating w.r.t.x]

So, the slope of line parallel to the X-axis.

\(\therefore (\frac{dy}{dx}) = 0\)

⇒ \(3x^2 - 12 = 0\)

⇒ \(x^2 = \frac{12}{3} = 4\)

\(\therefore x = \pm 2\)

For x = 2, y = \(2^3\) - 12 x 2 + 18 = 2

and for x = -2, y = \((-2)^3\) -12(-2) + 18 = 34

So, the points are (2, 2) and (-2, 34).

The equation of curve is \(y = e^{2x}\)

Since, it passes through the point (0, 1).

\(\therefore \frac{dy}{dx} = e^{2x}.2 = 2.e^{2x}\)

⇒ \((\frac{dy}{dx})_{x = 0} = 2e^{2 \times 0}\) = 2 = Slope of tangent to the curve

\(\therefore\) Equation of tangent is y -1 = 2(x - 0)

⇒ y = 2x + 1

Since, tangent to curve \(y = e^{2x}\) at the point (0, 1) meets X-axis i.e., y = 0.

\(\therefore\) 0 = 2x + 1 ⇒ x = \(-\frac{1}{2}\)

So, the required points is \((-\frac{1}{2}, 0)\)

Equation of curve is given is given by

x = \(t^2\) + 3t - 8 and y = 2\(t^2\) - 2t - 5

\(\therefore \frac{dx}{dt}\) = 2t + 3 and \(\frac{dy}{dx}\) = 4t - 2

⇒ \(\frac{dy}{dx}\) = \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t - 2}{2t + 3}\)  ...(i)

Since, the curve passes through the point (2, -1)

\(\therefore \) 2 = \(t^2\) + 3t - 8

and -1 = 2\(t^2\) - 2t - 5

⇒ \(t^2\) + 3t - 10 = 0

and 2\(t^2\) - 2t - 4 = 0

⇒ \(t^2\) + 5t - 2t - 10 = 0

⇒ 2\(t^2\) + 2t - 4t - 4 = 0

⇒ t(t + 5) - 2(t + 5) = 0

and 2t(t + 1) - 4(t + 1) = 0

⇒ (t - 2)(t + 5) = 0

and (2t - 4)(t + 1) = 0

⇒ t = 2, -5 and t = -1, 2

⇒ t = 2

\(\therefore \) Slope of tangent,

\((\frac{dy}{dx})_{t =2} = \frac{4 \times 2 - 2}{2 \times 2 + 3} = \frac{6}{7}\) [using Eq. (i)]

Equation of two curves are given by

\(x^3 - 3xy^2 + 2 = 0\)

and \(3x^2y - y^3 - 2 = 0\) [on differentiating w.r.t.x]

⇒ \(3x^2 - 3[x.2.y \frac{dy}{dx} + y^2.1] + 0 = 0\)

and \(3[x^2\frac{dy}{dx} + y.2x]-3y^2\frac{dy}{dx} - 0 = 0\)

⇒ \(3x.2y\frac{dy}{dx} + 3y^2 = 3x^2\)

and \(3y^2 \frac{dy}{dx} = 3y^2 \frac{dy}{dx} + 6xy\)

⇒ \(\frac{dy}{dx} = \frac{3x^2 - 3y^2}{6xy}\)

and \(\frac{dy}{dx} = \frac{6xy}{3y^2 - 3x^2}\)

⇒ \((\frac{dy}{dx}) = \frac{3(x^2 - y^2)}{6xy}\)

and \((\frac{dy}{dx}) = \frac{-6xy}{3(x^2 - y^2)}\)

⇒ \(m_1 = \frac{(x^2 - y^2)}{2xy}\)

and \(m_2 = \frac{-2xy}{x^2 - y^2}\)

\(\therefore m_1m_2 = \frac{x^2 - y^2}{2xy}.\frac{-(2xy)}{x^2 - y^2}\)

= -1

Hence, both the curves are intersecting at right angle i.e., making \(\frac{\pi}{2}\) with each other.

We have, f(x) = 2x + cos x

\(\therefore\) f'(x) = 2 + (-sin x) = 2 - sin x >0

Since, f'(x) > 0, \(\forall\) x \(\in\) R

Hence, f(x) is an increasing function.

In the interval \((0, \frac{\pi}{2})\) f(x) = cos x

⇒ f'(x) = -sin x

Which gives f'(x) < 0 in \((0, \frac{\pi}{2})\)

Hence, f(x) = cos x is decreasing in \((0, \frac{\pi}{2})\)

We have, f(x) = tan x - x

\(\therefore\) f'(x) = \(sec^2x - 1\)

⇒ f'(x) > 0, \(\forall\) x \(\in\) R

So, f(x) always increasing.

Let f(x) = \(x^2 - 8x + 17\)

\(\therefore\) f'(x) = 2x - 8

So, f'(x) = 0, gives x = 4

Now, f''(x) = 2 > 0, \(\forall\) x

So, x = 4 is the point of local minima.

\(\therefore\) Minimum value of f(x) at x = 4,

f(4) = 4 x 4 - 8 x 4 + 17 = 1

We have, \(f(x) = x^3 - 18x^2 + 96x\)

\(\therefore f'(x) = 3x^2 - 36x + 96\)

So, f'(x) = 0

Gives, \(3x^2 - 36x + 96 = 0 \)

⇒ \(3(x^2 - 12x + 32) = 0\)

⇒ (x - 8)(x - 4) = 0

⇒ x = 8, 4 \(\in\) [0, 9]

We shall now evaluate the value of f at these points and at the end points of the interval [0, 9] i.e., at x = 4 and x = 8 and at x = 0 and at x = 9

\(\therefore f(4) = 4^3 - 18.4^2 + 96.4\)

= 64 - 288 + 384 = 160

f(8) = \(8^3 - 18.8^2 + 96.8 = 128\)

f(9) = \(9^3 - 18.9^2 + 96.9\)

= 729 - 1458 + 864 = 135

and f(0) = \(0^3 - 18.0^2 + 96.0 = 0\)

Thus, we conclude that absolute minimum value of f on [0, 9] is 0 occurring at x = 0.

We have, f(x) = sinx.cos x \(= \frac{1}{2} sin 2x\)

\(\therefore\) f'(x) = \(\frac{1}{2}\).cos 2x.2 = cos 2x

Now, f'(x) = 0 ⇒ cos 2x = 0

⇒ cos 2x = cos \(\frac{\pi}{2}\) ⇒ x = \(\frac{\pi}{4}\)

Also, f''(x) = \(\frac{d}{dx}\)cos 2x = -sin2x.2 = -2 sin 2x

\(\therefore\) [f''(x)\(]_{atx = \pi/4}\) = -2. sin 2. \(\frac{\pi}{4}\) = - 2 sin \(\frac{\pi}{2}\) = -2 < 0

At \(\frac{\pi}{4}\), f(x) is maximum and \(\frac{\pi}{4}\) is point of maxima.

\(\therefore\) f(\(\frac{\pi}{4}\)) = \(\frac{1}{2}\). sin(2. \(\frac{\pi}{4}\))= \(\frac{1}{2}\)