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Application of Derivatives

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Short Answer Type

81.

Show that the normal at any point 0 to the curve
x = a cosө + a ө sinө. y = a sin ө – a ө cos ө
is at a constant distance from the origin.

78 Views

82.

Find the equation of the tangent and normal to the given curves at the points given:
y = x² at (0,0).

76 Views

83.

Find the equation of the tangent and normal to the given curves at the points given:
y² = 4ax at (0, 0).

89 Views

84.

Find the equation of the tangent and normal to the given curves at the points given:
y = x³ at (1, 1)

74 Views

85.

Find the equation of the tangent and normal to the given curves at the points given:
y = x² at (2, 8)

72 Views

Long Answer Type

86.

Find the equation of tangent and normal to the hyperbola $straight x squared over straight a squared minus straight y squared over straight b squared space equals space 1 space space space space at space the space point space space left parenthesis straight x subscript 0 comma space space space straight y subscript 0 right parenthesis$

80 Views

Short Answer Type

87. Find the equation of the normal at the point (am², am³) for the curve ay²= x³.

83 Views

88.
Find the equation of the tangent and the normal to the curve $straight x to the power of 2 over 3 end exponent plus straight y to the power of 2 over 3 end exponent space equals space 2 space space space at space space left parenthesis 1 comma space 1 right parenthesis.$

The equations of curve is
$straight x to the power of 2 over 3 end exponent plus straight y to the power of 2 over 3 end exponent space equals space 2$

Differentiating both sides w.r.t. x, we get,

$2 over 3 straight x to the power of minus to the power of 1 third end exponent end exponent space plus space 2 over 3 straight y to the power of negative 1 third end exponent dy over dx space equals space 0 space space space space space space space space or space space space space space space straight y to the power of negative 1 third end exponent dy over dx space equals negative straight x to the power of negative 1 third end exponent therefore space space space space space space space dy over dx space equals space minus straight x to the power of negative begin display style 1 third end style end exponent over straight y to the power of negative begin display style 1 third end style end exponent space equals space minus straight y to the power of begin display style 1 third end style end exponent over straight x to the power of begin display style 1 third end style end exponent space equals space minus open parentheses straight y over straight x close parentheses to the power of 1 third end exponent At space left parenthesis 1 comma space 1 right parenthesis comma space slope space of space tangent space equals space minus open parentheses 1 over 1 close parentheses to the power of 1 third end exponent space equals space minus 1 therefore space space space space space space space space equation space of space tangent space at space left parenthesis 1 comma space 1 right parenthesis space is space space space space space space space space space space space space space space straight y minus 1 space equals space minus left parenthesis straight x minus 1 right parenthesis comma space space space or space space space straight y minus 1 space equals space minus straight x plus 1 space space space or space space space space straight x plus straight y minus 2 space equals space 0 At space left parenthesis 1 comma space 1 right parenthesis comma space slope space of space normal space space equals space minus fraction numerator 1 over denominator negative 1 end fraction space equals space 1 therefore space space space space space space space equation space of space normal space at space left parenthesis 1 comma space 1 right parenthesis space is space space space space space space space space space space space space space space space space space space space space space space space space space straight y minus 1 space equals space 1 left parenthesis straight x minus 1 right parenthesis space space space space space or space space space space space space space space straight y minus 1 space equals space straight x minus 1 space space space space space space or space space space straight x minus straight y space equals space 0$

84 Views

89. Find the equation of the tangent and normal to the curve y² = 4ax at (at², 2at)

78 Views

Long Answer Type

90.

Find the equation of the tangent and normal to the curve
$straight x equals space 1 minus cos space straight theta comma space space space space straight y space equals space straight theta space minus sin space straight theta space space space at space straight theta space equals space straight pi over 4$