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1. Using differentials, find the approximate value of each of the following up to 3 places of decimal.

(iii)  \sqrt {0.6}
  

Lets suppose  and let x = 1 and  Then, Now, we cam say that   is approximately equals to dy Now, Hence,  is approximately equal to 0.8         

1. Using differentials, find the approximate value of each of the following up to 3 places of decimal. 

(ii)\sqrt { 49.5 }

Lets suppose  and let x = 49 and  Then, Now, we can say that   is approximately equal to dy Now, Hence,  is approximately equal to 7.035         

1. Using differentials, find the approximate value of each of the following up to 3
places of decimal. (i)  \sqrt {25.3 }

Lets suppose  and let x = 25 and  Then, Now, we can say that   is approximate equals to dy Now, Hence,  is approximately equals to 5.03          

27) The line y = x+1 is a tangent to the curve y^2 = 4 x at the point
(A) (1, 2)                         (B) (2, 1)                             (C) (1, – 2)                                        (D) (– 1, 2)

The slope of the given line  is 1 given curve equation is   If the line is tangent to the given curve than the slope of the tangent is equal to the slope of the curve The slope of tangent =  Now, when y = 2,  Hence, the coordinates are (1,2) Hence, (A) is the correct answer  

26) The slope of the normal to the curvey = 2x ^2 + 3 \sin x \: \: at \: \: x = 0is
(A) 3                                       (B) 1/3                               (C) –3                                           (D) -1/3

Equation of the given curve is  Slope of tangent =  at x = 0 Now, we know that Hence, (D) is the correct option 

25) Find the equation of the tangent to the curve y = \sqrt{3x-2} which is parallel to the line 4x - 2y + 5 = 0 .

Parellel to line  means the slope of tangent and slope of line is equal  We know that the equation of line is y = mx + c on comparing with the given equation we get the slope of line m = 2 and c = 5/2 Now, we know that the  slope of the tangent at a given point to given curve is given by  Given the equation of curve is   Now, when  ,      but y cannot be -ve so we take only positive...

24) Find the equations of the tangent and normal to the hyperbola
\frac{x^2 }{a^2} - \frac{y^2 }{b^2 }= 1 at the point (x_0 , y_0 )

Given equation is  Now ,we know that  slope of tangent =  at point  equation of tangent at point  with slope  Now, divide  both sides by                             Hence, the equation of tangent is                                                     We know that equation of normal  at the point  with slope  

23) Prove that the curves x = y^2 and xy = k cut at right angles* if \: \: 8k ^ 2 = 1.

Let suppose, Curve  and xy = k cut at the right  angle then the slope of their tangent also cut  at the right angle means,                                                                    -(i) Now these values in equation (i)                               Hence proved

22) Find the equations of the tangent and normal to the parabola y ^2 = 4 ax at the point (at ^2, 2at).

Equation of the given curve is Slope of tangent =  at point  Now, the equation of tangent with point  and slope  is We know that Now, the equation of at point    with slope -t   Hence, the equations of the tangent and normal to the parabola  at the point  are

21) Find the equation of the normals to the curvey = x^3 + 2x + 6 which are parallel
to the line x + 14y + 4 = 0.

Equation of given curve is Parellel to line  means slope of normal and line is equal We know that, equation of line y= mx + c on comparing it with  our given equation. we get, Slope of tangent =  We know that Now, when x = 2,   and  When x = -2 ,  Hence, the coordinates are (2,18) and (-2,-6) Now, the equation of at point (2,18) with slope  Similarly,  the equation of at point...

20) Find the equation of the normal at the point ( am^2 , am^3 ) for the curve ay ^2 = x ^3.

Given equation of curve is  Slope of tangent   at point  Now, we know that  equation of normal at point  and with slope  Hence, the equation of normal is 

19) Find the points on the curve x^2 + y^2 - 2x - 3 = 0 at which the tangents are parallel
to the x-axis.

parellel to x-axis means slope is 0 Given equation of curve is Slope of  tangent =   When x = 1 ,                          Hence, the coordinates are (1,2) and (1,-2)

18) For the curve y = 4x ^ 3 - 2x ^5, find all the points at which the tangent passes
through the origin.

Tangent passes through origin so, (x,y) = (0,0) Given equtaion of curve is  Slope of tangent = Now, equation of tangent is  at (0,0)  Y =  0 and X = 0 and we have  Now, when x = 0,    when x = 1 ,  when x= -1 ,   Hence, the coordinates are (0,0) , (1,2) and (-1,-2)

17) Find the points on the curve y = x ^3 at which the slope of the tangent is equal to the y-coordinate of the point.

Given equation of curve is   Slope of tangent =  it is given that  the slope of the tangent is equal to the y-coordinate of the point We have   So, when x = 0 , y = 0 and when x = 3 ,  Hence, the coordinates are (3,27) and (0,0)

16) Show that the tangents to the curve y = 7x^3 + 11 at the points where x = 2 and x = – 2 are parallel.

Slope of tangent =   When x = 2 When  x = -2 Slope is equal when x= 2 and x = - 2  Hence, we can say that both the tangents to curve  is parallel

15) Find the equation of the tangent line to the curve y = x^2 -2x +7 which is (b) perpendicular to the line 5y - 15x = 13.

Perpendicular to line    means  We know that the equation of the line is y = mx + c on comparing with the given equation we get the slope of line m = 3 and c = 13/5 Now, we know that the  slope of the tangent at a given point to given curve is given by  Given the equation of curve is   Now, when  ,  Hence, the coordinates are  Now, the equation of tangent passing through (2,7) and...

15) Find the equation of the tangent line to the curve y = x^2 - 2x +7 which is  (a) parallel to the line 2x - y + 9 = 0

Parellel to line  means slope of tangent and slope of line is equal  We know that the equation of line is y = mx + c on comparing with the given equation we get slope of line m = 2 and c = 9 Now, we know that the  slope of tangent at a given point to given curve is given by  Given equation of curve is   Now, when x = 2 ,  Hence, the coordinates are (2,7) Now, equation of tangent...

14) Find the equations of the tangent and normal to the given curves at the indicated points:

e) x = \cos t , y = \sin t \: \: at \: \: t = \pi /4

We know that Slope of the tangent at a point on the given curve is given  by   Given the equation of the curve Now,             and            Now,  Hence slope of the tangent is -1 Now we know that, Now, the equation of the tangent at the point  with slope = -1 is           and             equation of the tangent at   i.e.  is Similarly, the equation of normal at   with slope =...

14) Find the equations of the tangent and normal to the given curves at the indicated points

d)  y = x^2\: \: at\: \: (0, 0)

We know that Slope of the tangent at a point on the given curve is given  by   Given the equation of the curve at point (0,0) Hence slope of tangent is 0 Now we know that, Now, equation of tangent at point (0,0) with slope = 0 is y = 0 Similarly, equation of normal at point (0,0) with slope =  is

14) Find the equations of the tangent and normal to the given curves at the indicated
points:

c)  y = x^3\: \: at \: \: (1, 1)

We know that Slope of the tangent at a point on the given curve is given  by   Given the equation of the curve at point (1,1) Hence slope of tangent is 3 Now we know that, Now, equation of tangent at point (1,1) with slope = 3 is equation of tangent is Similarly, equation of normal at point (1,1) with slope = -1/3 is y = mx + c equation of normal is
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