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Application of Derivatives

Name: Zigya - For The Curious Learner
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Short Answer Type

401.

Find the equation of tangents to the curve y= x³+2x-4, which are perpendicular to line x+14y+3 =0.

2336 Views

Long Answer Type

402.

Prove space that space straight y equals fraction numerator 4 space sin space straight theta over denominator 2 space plus cos space straight theta end fraction space minus straight theta space is space an space increasing space function space of space straight theta space on space open square brackets 0 comma straight pi over 2 close square brackets

1368 Views

403.

Show that semi-vertical angle of a cone of maximum volume and given slant height is cos^-1 $fraction numerator 1 over denominator square root of 3 end fraction$ .

1713 Views

404.

Find the absolute maximum and absolute minimum values of the function f given by
$straight f left parenthesis straight x right parenthesis space equals sin squared straight x minus cosx comma space straight x space element of space left parenthesis 0 comma space straight pi right parenthesis$

619 Views

Short Answer Type

405.

Find the value(s) of x for which y = $open square brackets straight x left parenthesis straight x minus 2 right parenthesis close square brackets squared$ is an increasing function.

237 Views

406.

Find the equations of the tangent and normal to the curve $straight x squared over straight a squared minus straight y squared over straight b squared equals 1 space at space the space point space open parentheses square root of 2 straight a comma space straight b close parentheses.$

162 Views

Long Answer Type

407.
If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum when the angle between them is $60 degree.$

Let ABC be the right angle triangle with base b and hypotenuse h.
Given that b+ h = k
Let A be the area of the right triangle.

$straight A equals 1 half cross times straight b cross times square root of straight h squared minus straight b squared end root rightwards double arrow space space straight A squared space equals space 1 fourth straight b squared left parenthesis straight h squared minus straight b squared right parenthesis rightwards double arrow space straight A squared equals space space straight b squared over 4 open parentheses open parentheses straight k minus straight b close parentheses squared minus straight b squared close parentheses space space space space space space open square brackets because space straight h space equals space straight k minus straight b close square brackets rightwards double arrow straight A squared space equals space straight b squared over 4 open parentheses straight k squared plus straight b squared minus 2 kb minus straight b squared close parentheses rightwards double arrow straight A squared space equals space straight b squared over 4 left parenthesis straight k squared minus 2 kb right parenthesis rightwards double arrow straight A squared equals space fraction numerator straight b squared straight k squared minus 2 kb cubed over denominator 4 end fraction$
Differentiating the above function with respect to be, we have

$2 straight A dA over db equals fraction numerator 2 bk squared minus 6 kb squared over denominator 4 end fraction space... left parenthesis 1 right parenthesis rightwards double arrow space space dA over db equals fraction numerator bk squared minus 3 kb squared over denominator 2 straight A end fraction$
For the area to be maximum, we have

$dA over db equals 0$
$rightwards double arrow space bk squared minus 3 kb squared space equals 0 rightwards double arrow space bk space equals space 3 straight b squared rightwards double arrow space straight b space equals space straight k over 3$
Again differentiating the function in equation (1), with respect to b, we have
$2 open parentheses dA over db close parentheses squared plus 2 straight A fraction numerator straight d squared straight A over denominator db squared end fraction equals fraction numerator 2 straight k squared minus 12 kb over denominator 4 end fraction space... left parenthesis 2 right parenthesis$
Now substituting $dA over db equals 0 space and space straight b space equals space straight k over 3$ in equation (2), we have
$2 straight A fraction numerator straight d squared straight A over denominator db squared end fraction space equals space fraction numerator 2 straight k squared minus 12 straight k open parentheses begin display style straight k over 3 end style close parentheses over denominator 4 end fraction space space space rightwards double arrow 2 straight A fraction numerator straight d squared straight A over denominator db squared end fraction space equals space fraction numerator 6 straight k squared minus 12 straight k squared over denominator 12 end fraction space space rightwards double arrow space 2 straight a fraction numerator straight d squared straight A over denominator db squared end fraction equals negative straight k squared over 2 space space space rightwards double arrow space fraction numerator straight d squared straight A over denominator db squared end fraction equals negative fraction numerator straight k squared over denominator 4 straight A end fraction less than 0$
Thus area is maximum at $straight b equals straight k over 3.$
Now, $straight h equals straight k minus straight k over 3 equals fraction numerator 2 straight k over denominator 3 end fraction$
Let $straight theta$ be the angle between the base of the triangle and the hypotenuse of the right angle.

$Thus comma space cosθ equals straight b over straight h equals fraction numerator begin display style straight k over 3 end style over denominator begin display style fraction numerator 2 straight k over denominator 3 end fraction end style end fraction equals 1 half rightwards double arrow space straight theta equals space cos to the power of negative 1 end exponent open parentheses 1 half close parentheses equals space straight pi over 3$

162 Views

408.

Show that the height of the cylinder of maximum volume, which can be inscribed in a sphere of radius R is Also find the maximum volume.

490 Views

409.

Find the equation of the normal at a point on the curve x² = 4y which passes through the point (1, 2). Also, find the equation of the corresponding tangent.

336 Views

Short Answer Type

410.

The volume of a sphere is increasing at the rate of 3 cubic centimetres per second. Find the rate of increase of its surface area, when the radius is 2 cm.

3950 Views