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toni_malvina toni_malvina
wrote...
12 years ago
Use implicit differentiation to find the following.

8x^2+2y^2=202, (5, -1)
(a) The slope of the tangent line at the indicated point on the graph.
m =

(b) The equation of the tangent line at the indicated point on the graph.
y =
2
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juicylucy Author
wrote...
#1
12 years ago
16x +4y dy/dx =0     then, dy/dx = -4x/y  at (5,-1)  dy/dx =m= 20

equation is ,  y+1 =20(x-5)  y= 20x -101
wrote...
#2
12 years ago
16x + 4y dy / dx = 0
4y dy / dx = -16x
dy / dx = -16x / (4y) = -4x / y

when x = 5 and y = -1, then dy/dx = -4(5) / (-1) = 20 = slope of tangent at (5 , -1)

the line with slope 20 passing through (5 , -1) looks like: -1 = 20(5) + b
-101 = b

thus, y = 20x - 101 is the tangent line through (5 , -1)
wrote...
#3
12 years ago
use the power rule

16 x + 4y dy = 0

now solve for dy

dy = -16x / 4y

plug in the values

dy = -16(5) / 4(-1)

dy = 20

that is your slope.  Now, use the point slope form to make the line.

y = m( x - x1) + y1

y = 20( x - 5) - 1

y = 20x -101
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