Can i get some help with these problems, the porffessor skipped around a lot and only did 2 questions in this section that were not remotely like this at all. So if someone could explain step by step how to solve this problem it would be greatly appreciated 1) when y=2x^2+8x+4 and dx/dt=13 when x=6 2) where y=√(x)...
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answers:
Philip say: 1) y = 2x^2 +8x +4. (dy/dx) = 4x+8 = 4(x+2). (dy/dt) = (dy/dx)(dx/dt). When x = 6, (dx/dt) = 13 and
(dy/dt) = 4(x+2)(dx/dt) = 4(6+2)13 = 32*13 = 416.
2) y = rt(x) -3 = x^(1/2) -3. (dy/dx) = (1/2)x^(-1/2). (dy/dt) = (dy/dx)(dx/dt). When x = 9, (dx/dt) = 2 and
(dy/dt) = (1/2)[x^(-1/2)](dx/dt) = [(1/2)9^(-1/2)](2) = (1/3).
3) x^2 +xy -y^2 = 0. Put (dy/dx) = y'. Differentiating [x^2 +xy -y^2 = 0] gives 2x +(y +xy') -2yy' = 0, ie.,
y'(2y -x) = (2x +y), ie., (dy/dx) = [(2x+y)/(2y-x)]. (dy/dt) = (dy/dx)(dx/dt). when x = 4, (dx/dt) = 2 and
(dy/dt) = [(2x+y)/(2y-x)](2). Now, when x = 4, x^2 + xy -y^2 = 0. ie., (4)^2 +(4)y -y^2 = 0, ie.,
y^2 -4y -16 = 0 and 2y = 4(+/-)rt[(4)^2 -4(1)(-16)] = 4(+/-)rt80 = 4[1(+/-)rt5], ie., y = 2[1(+/-)rt5]. So @x = 4, (dy/dt) = 2{8+2[1(+/-)rt5]}/{4[1(+/-)rt5]-4} = 4[5(+/-)rt5]/[(+/-)4rt5] = (+/-)(1+rt5).
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Amy say: Use the chain rule.
dy/dt = dy/dx * dx/dt
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poornakumar b say: Since you introduced t as the independent variable, it can be written as dx/dt=2 when x=9
dy/dt = (dy/dx)(dx/dt)
1) y=2x² +8x+4.
dy/dx = 4x+8,
dy/dt = (dy/dx)(dx/dt) = 13 (4x+8)
= 13(4.6 + 8) = 416 ... for x=6
2) y=√(x) -3
dy/dx = d/dt[ₓ½ -3] = ₓ-½ = 1/√(x)
dy/dt = (dy/dx)(dx/dt) = 2/√(x)
= 2/√(9) = 2/3 ... for x=9
3) x²+xy-y² = 4 and dx/dt=2 when x=4
2x + 1(dy/dx) + 2y(dy/dx) = 0,
(dy/dx)[1 -2y] = -2x,
(dy/dx)[2y -1] = 2x,
(dy/dx) = 2x/[2y -1].