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Application of Derivatives

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Long Answer Type

431.

Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of t heradius of the base. How fast is the sand cone increasing when the height is 4 cm?

432.

Find the points on the curve x² + y² – 2x – 3= 0 at whichthe tangents are parallel to x-axis.

433.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Let the rectangle of length l and breadth b be inscribed in circle of radius a.

Then, the diagonal of the rectangle passes through the centre and is of length 2a cm.

Now, by applying the Pythagoras Theorem, we have:

( 2a )² = l² + b²

$\Rightarrow b2= 4 a2-l2$

$\Rightarrow b = \sqrt{4 a^{2} - l^{2}} ∴ Area of rectangle, A = l \sqrt{4 a^{2} - l^{2}} ∴ \frac{dA}{dl} = \sqrt{4 a^{2} - l^{2}} + l \frac{1}{2 \sqrt{4 a^{2} - l^{2}}} (- 2 l) = \sqrt{4 a^{2} - l^{2}} - \frac{l^{2}}{\sqrt{4 a^{2} - l^{2}}} = \frac{4 a^{2} - 2 l^{2}}{\sqrt{4 a^{2} - l^{2}}} \frac{{dA}^{2}}{{dl}^{2}} = \frac{\sqrt{4 a^{2} - l^{2}} (- 4 l) - (4 a^{2} - 2 l^{2}) \frac{(- 2 l)}{2 \sqrt{4 a^{2} - l^{2}}}}{(4 a^{2} - l^{2})} = \frac{(4 a^{2} - l^{2}) (- 4 l) + l (4 a^{2} - 2 l^{2})}{{(4 a^{2} - l^{2})}^{\frac{3}{2}}} = \frac{- 12 a^{2} l + 2 l^{3}}{{(4 a^{2} - l^{2})}^{\frac{3}{2}}} = \frac{- 2 l (6 a^{2} - l^{2})}{{(4 a^{2} - l^{2})}^{\frac{3}{2}}}$

$Now, \frac{dA}{d l} = 0 gives 4 a^{2} = 2 l^{2} \Rightarrow l = \sqrt{2 a} \Rightarrow b = \sqrt{4 a^{2} - 2 a^{2}} = \sqrt{2 a^{2}} = \sqrt{2} a When l = \sqrt{2} a, \frac{{dA}^{2}}{d l^{2}} = \frac{- 2 (\sqrt{2} a) (6 a^{2} - 2 a^{2})}{2 \sqrt{2} a^{3}} = \frac{- 8 \sqrt{2} a^{3}}{2 \sqrt{2} a^{3}} = - 4 < 0$

$∴$ Thus, frrom the second derivative test, when l = $\sqrt{2} a$ , the area of the rectangle is maximum.

Since l = b = $\sqrt{2} a$ , the rectangle is square.

Hence, of all the rectangles inscribed in the given circle, the square has the maximum area.

434.

Find the point on the curve y = x³ – 11x + 5 at which the equation of tangent is y = x – 11.

435.

Using differentials, find the approximate value of $\sqrt{49.5} .$

436.

Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal to the diameter of its base.

Multiple Choice Questions

437.

On the ellipse 4x² + 9y² = 1, the points at which the tangents are parallel to the line 8x = 9y, are

$(\frac{2}{5}, \frac{1}{5})$
$(- \frac{2}{5}, \frac{1}{5})$
$(- \frac{2}{5}, - \frac{1}{5})$
$(\frac{2}{5}, - \frac{1}{5})$

438.

A container s the shape of an inverted cone. Its height is 6 m and radius is 4m at the top. If it is filled with water at the rate of 3m/min then the rate of change of height of water(in mt/min) when the water level is 3 m is

$\frac{3}{4 π}$
$\frac{2}{9 π}$
$16 π$
$2 π$

439.

$If α, β, γ are the lengths of the tangents from the vertices of a triangle to its incircle . Then$

$α + β + γ = \frac{1}{r^{2}} (αβγ)$
$α + β + γ = \frac{1}{r} (αβγ)$
$\frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = r (αβγ)$
$α^{2} + β^{2} + γ^{2} = \frac{2}{r} (αβγ)$

440.

If a cylindrical vessel of given volume V with no lid on the top is to be made from a sheet of metal, then the radius (r) and height(h) of the vessel so that the metal sheet used is minimum is

$r = \sqrt[3]{\frac{π}{V}}, h = \sqrt[3]{\frac{π}{V}}$
$r = \sqrt{πV}, h = \sqrt{πV}$
$r = \sqrt[3]{\frac{V}{π}}, h = \sqrt[3]{\frac{V}{π}}$
$r = \sqrt{\frac{V}{π}}, h = \sqrt{\frac{V}{π}}$