Find dy/dx , in terms of t , of the following parametric function.
x = 2t / (4+t)
y = sqrt[( t^2) +1] / t
((please help me......thz a lot))
HELP...Differentiation
2007-10-04 6:53 am
回答 (3)
2007-10-04 7:09 am
參考: My Maths knowledge
2007-10-04 7:12 am
dx/dt = [(4+t)(2) + 2t(1)] / (4+t)^2 = (4t+8)/(4+t)^2
dy/dt = [t*1/2 (t^2+1)^(-1/2) (2t) - (t^2+1)^(1/2)] /t^2
dy/dt = [t^2/sqrt(t^2+1) - sqrt(t^2+1)] /t^2 = t^2 -t^2-1 /(t^2 * sqrt(t^2+1))
=-1/t^2 sqrt(t^2+1)
dy/dx = (dy/dt) / (dx/dt) = -(4+t)^2/[t^2(4t+8)sqrt(t^2+1)]
dy/dt = [t*1/2 (t^2+1)^(-1/2) (2t) - (t^2+1)^(1/2)] /t^2
dy/dt = [t^2/sqrt(t^2+1) - sqrt(t^2+1)] /t^2 = t^2 -t^2-1 /(t^2 * sqrt(t^2+1))
=-1/t^2 sqrt(t^2+1)
dy/dx = (dy/dt) / (dx/dt) = -(4+t)^2/[t^2(4t+8)sqrt(t^2+1)]
2007-10-04 7:09 am
x = 2t/(4+t)
dx/dt
= [(2t)' (4+t) - (2t)(4+t)']/(4+t)^2
= [(2)(4+t)-(2t)(1)]/(4+t)^2
= 8/(4+t)^2
y = sqrt[(t^2)+1]/t
dy/dt
= {{sqrt[(t^2)+1]}' (t) - sqrt[(t^2)+1] (t)'}/t^2
= .....
= -1/{2t sqrt[(t^2)+1]}
dy/dx = (dy/dt) / (dx/dt)
...... hope you can do it !!!
dx/dt
= [(2t)' (4+t) - (2t)(4+t)']/(4+t)^2
= [(2)(4+t)-(2t)(1)]/(4+t)^2
= 8/(4+t)^2
y = sqrt[(t^2)+1]/t
dy/dt
= {{sqrt[(t^2)+1]}' (t) - sqrt[(t^2)+1] (t)'}/t^2
= .....
= -1/{2t sqrt[(t^2)+1]}
dy/dx = (dy/dt) / (dx/dt)
...... hope you can do it !!!
收錄日期: 2021-04-13 18:12:17
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