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Find equations of the tangent line and normal line to the curve at the given point?

+1 vote

47. y = sqrt (1+4sinx) , (0,1)

48. x^2 + 4xy + y^2 = 13, (2,1)

please show all steps so i can understand

asked Mar 20, 2013 in CALCULUS by linda Scholar

4 Answers

+1 vote

47.

The curve y = √(1+ 4sinx)

The point of the curve is (0,1)

The slope of the curve m = dy / dx

But y = √(1+ 4sinx)

Diferenciate with respective x

dy / dx = d / dx√(1 + 4sinx)

d / dx(sinx) = cosx

dy / dx = 1 / 2√(1 + 4sinx)(0 + 4cosx)

The slope of the curve at point (0,1) : m = dy / dx = 1 / 2√(1 + 4sin0)(0 + cos0)

m = dy / dx = 1 / 2√(1 + 0)(0+ 1)

m = 1/ 2 √(1)(1)

m = 1 / 2

The slope of the curve m and at the point (x1,y1) then  the equation of the tangent is y - y1 = m(x - x1)

Substitute m = 1 / 2 and x1 = 0 , y1 = 1

Tangent line is y - 1 = 1 / 2(x - 0)

y = x / 2 + 1.

The normal line slope is -1 / m = -1 / 1/2

slope =-2 and point (0,1)

The equation is y - 1 = -2(x - 0)

There fore y = -2x + 1

The equation of the normal line : y = -2x + 1.

 

 

answered Mar 23, 2013 by diane Scholar
0 votes

48.

 The curve x2 + 4xy + y2 = 13

The point of the curve is (2,1)

The slope of the curve m = dy / dx

Diferenciate with respective x to the curve

Recall : Diferenciate formula deriavative of xn  = n xn-1 and derivative of uv = uv1 +vu1

2x + 4[x (dy / dx) + y(1)] + 2y dy / dx = 2x + 4[x (dy / dx) + y] + 2y dy / dx = o

2x + 4xdy /dx + 4xy + 2ydy /dx = 0

2x + 4xdy / dx + 2ydy / dx  + 4xy = 0

2x + 4xy + 2dy / dx (2x + y) = 0

Subtract 2x + 4xy from each side

2dy / dx (2x + y) = -(2x + 4xy)

Take out common term 2x

2dy / dx (2x + y) = -2x (1 + 2y)

Divide each side by 2

dy /dx (2x + y) = -(1 + 2y)

Divide each side by 2x + y

dy /dx = -(1+2y) / (2x + y)

dy / dx at point (2,1)

Substitute x = 2 and y = 1 in the slope (m) dy /dx

The slope m = -(1 + 2(1)) / (2(2) + 1)

Simplify

m = -3 / 5

The slope of the given curve is m = -3 / 5 and at point (2,1)

slope m and point (x1,y1) then

The equation of the tangent line : y - y1 = m( x - x1)

Substitute m = -3 / 5 and x1 = 2 , y1= 1 in the equation of the tangent line y -1 = -3/5(x - 2)

y - 1 = -3/5x +6/5

Add 1 to each side

y = -3/5x + 6 / 5 + 1

Simplify

y =-3/5x + 11 / 5

Multiply each side by 5

There fore 5y = -3x + 11

The equation of the tangent line : y = -3/5x + 11/5 or 3x + 5y = 11

The product slopes of the tangent and normal line is -1.

There fore the normal slope = -1 / tangent slope =-1 / m

Substitute m = -3 / 5

The normal slope = -1 / -3/5 =-5 / -3 = 5 / 3 and at point (2,1)

Substitute slope =5/3, x1 =2 and y1 = 1

The normal line equation y - 1 = 5/3( x - 2)

Simplify

y - 1 = 5/3x - 10/3

Add 1 to each side

y = 5/3x -10/3 +1

Simplify

y = 5/3x -7/3

Multiply each side by 3

3y =5x - 7

There fore the normal line equation : 5x - 3y =7

 

answered Mar 23, 2013 by diane Scholar
this is wrong, you have one more x than you should when you derive.

Please check the updated answer.

0 votes

 

47.

The curve y = √(1+ 4sinx)

The point of the curve is (0,1)

The slope of the curve m = dy / dx

But y = √(1+ 4sinx)

Diferenciate with respective x

dy / dx = d / dx√(1 + 4sinx)

d / dx(sinx) = cosx

dy / dx = 1 / 2√(1 + 4sinx)(0 + 4cosx)

The slope of the curve at point (0,1) : m = dy / dx = 1 / 2√(1 + 4sin0)(0 + 4cos0)

m = dy / dx =  1/2(0+4)

m = dy / dx = 1/2*4 = 2

m = 2

The slope of the curve at the point (x1,y1) then equation of the tangent is y - y1 = m( x- x1)

Substitute m = 2 and x1 =0 , y1 =1

The equation of tangent line is y - 1 = 2(x - 0)

y - 1 = 2x

The normal line slope is -1/m = -1/2

Slope = -1/2 and point ( 0,1)

The equation is y - 1 =-1/2(x -0)

y-1 =-1/2(x)

2(y - 1) = -x

Therefore x + 2y - 2 = 0

The equation of normal line is x + 2y - 2 =0

 

answered May 13, 2013 by jeevitha Novice
0 votes

(48)

Step 1:

The curve equation is image and the point is image.

Slope of the function is derivative of the function at that point.

image

Apply derivative on each side with respect to image.

image

image

image

image

image

image

image.

Find the slope of the tangent line at image.

image

image

image.

Step 2:

Point slope form of line equation is image.

Substitute image and image.

image

image

image

image

The tangent line equation is image.

Step 3:

The normal line is perpendicular to the tangent line.

The equation of normal line at image is image, Where image is slope of the tangent line.

Substitute image and image.

image

image

image

image

image

The normal line equation is image.

Solution:

The tangent line equation is image.

The normal line equation is image.

answered Dec 15, 2015 by Sammi Mentor
edited Dec 15, 2015 by Sammi

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