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Mathematics: Post your doubts here!

X

xyz!

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hiee..jst a small doubt i hav! when v plot a complex no. on an Argand's diagram.. r v supposed to draw lines frm dat point to the x-axis and all...lyk v draw ven finding out mod or arg of that complex no.?
 
whitecorp

whitecorp

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salvatore said:
I need help in the following differentiation question:
Find the equation of the tangent to the curve y = x^2, which is parallel to the line y = x.

Please reply with a detailed solution..
Click to expand...

gradient function of curve is dy/dx = 2x

Since this particular tangent is parallel to the line y=x, then it must have a gradient value of 1 (compare this with the structure y=mx+c)

Hence,we have 2x= 1 =====> x=0.5

When x =0.5, y= 0.5^2 =0.25

Therefore, the equation of the required tangent parallel to the line y=x is

y-0.25 =(1) (x- 0.5) ======> y= x- 0.25 (shown)


Hope this helps. Peace.
 
whitecorp

whitecorp

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snowbrood said:
can anyone solve this SinA=2sinB and CosA+2CosB=2
Click to expand...

sinA =2sinB -------(1)
cosA +2 cosB =2 =======> cosA =2(1-cosB) -----------(2)
Squaring (1), we have sin^2 A = 4 sin^2 B ----------(3)
Squaring (2), we have cos^2 A= 4(1-cosB)^2 ---------(4)

(3)+(4): 1= 4 sin^2 B + 4 (1-cosB)^2
1= 4 sin^2 B + 4 - 8cos B + 4cos^2 B
1 = 4 ( sin^2 B + cos^2 B) +4 -8cosB
1= 4 +4 -8cosB
8 cosB =7 =====> B = arc cos (7/8) =28.96 deg (shown) and A=75.52 deg (shown)

Hope this helps. Peace.
 
snowbrood

snowbrood

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whitecorp said:
sinA =2sinB -------(1)
cosA +2 cosB =2 =======> cosA =2(1-cosB) -----------(2)
Squaring (1), we have sin^2 A = 4 sin^2 B ----------(3)
Squaring (2), we have cos^2 A= 4(1-cosB)^2 ---------(4)

(3)+(4): 1= 4 sin^2 B + 4 (1-cosB)^2
1= 4 sin^2 B + 4 - 8cos B + 4cos^2 B
1 = 4 ( sin^2 B + cos^2 B) +4 -8cosB
1= 4 +4 -8cosB
8 cosB =7 =====> B = arc cos (7/8) =28.96 deg (shown) and A=75.52 deg (shown)

Hope this helps. Peace.
Click to expand...
thanks a ton sir
 
salvatore

salvatore

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whitecorp said:
gradient function of curve is dy/dx = 2x

Since this particular tangent is parallel to the line y=x, then it must have a gradient value of 1 (compare this with the structure y=mx+c)

Hence,we have 2x= 1 =====> x=0.5

When x =0.5, y= 0.5^2 =0.25

Therefore, the equation of the required tangent parallel to the line y=x is

y-0.25 =(1) (x- 0.5) ======> y= x- 0.25 (shown)


Hope this helps. Peace.
Click to expand...

Thanks a lot man..
 
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