- #1
GregA
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Implicitly Defined Parametrization
I'm having difficulties with the following question, and having checked through my working several times I just can't find a problem...problem is, so far in the book implicit and parametric differentiation have been covered independently of each other and this question has just been thrown into the mix.
Find the slope when t = 0:
[tex] x + 2x^{\frac {3}{2}} = t^2 + t[/tex], [tex] y \sqrt{t+1} + 2t \sqrt{y} = 4 [/tex]
I can't see any other way to work with this other than to differentiate both equations w.r.t.t, find dy/dx by dividing dy/dt by dx/dt, and then plug in my numbers at the end.
dx/dt: [tex] \frac{dx}{dt} +3 \sqrt{x} \frac{dx}{dt} = 2t + 1[/tex]
[tex] \frac{dx}{dt} =\frac{2t + 1}{3 \sqrt{x}+1}[/tex]
dy/dt: [tex] \sqrt{t+1} \frac{dy}{dt} + \frac {y}{2\sqrt{t+1}} + \frac{t}{\sqrt{y}}\frac{dy}{dt}+ 2\sqrt{y} = 0[/tex]
just going to assign dummy variables:[tex] a = \sqrt{y}, b = \sqrt{t+1} [/tex] whilst I rearrange things...
[tex] \frac{dy}{dt}(b + \frac{t}{a}) = -(\frac {y}{2b} +2a)[/tex]
[tex] \frac{dy}{dt}(\frac{ba +t}{a}) = -(\frac {y +4ab}{2b})[/tex]
[tex] \frac{dy}{dt} = -(\frac {a(y +4ab)}{2b(ba+t)}) = -(\frac {\sqrt{y}(y + 4\sqrt{y}\sqrt{t+1})}{2\sqrt{t+1}(\sqrt{y}\sqrt{t+1}+t)})[/tex]
dy/dx =[tex] -\frac {(3 \sqrt{x}+1)(\sqrt{y}(y + 4\sqrt{y}\sqrt{t+1}))} {(2t + 1)(2\sqrt{t+1}(\sqrt{y}\sqrt{t+1}+t))}[/tex]
By using t = 0 and finding x and y in the original equations I get x = 1/4 and y = 4
[tex] -\frac{(\frac{3}{2}+1)(8 + 16)}{4} = -15 [/tex]
The answer I'm looking for however is -6. Is there some screw up with my working somewhere or am I using the wrong method?
I cannot graph any similar (but simpler) curves with Maxima to find the correct method neither because there doesn't seem to be a way to graph implicit expressions...nor will it differentiate one
I'm having difficulties with the following question, and having checked through my working several times I just can't find a problem...problem is, so far in the book implicit and parametric differentiation have been covered independently of each other and this question has just been thrown into the mix.
Find the slope when t = 0:
[tex] x + 2x^{\frac {3}{2}} = t^2 + t[/tex], [tex] y \sqrt{t+1} + 2t \sqrt{y} = 4 [/tex]
I can't see any other way to work with this other than to differentiate both equations w.r.t.t, find dy/dx by dividing dy/dt by dx/dt, and then plug in my numbers at the end.
dx/dt: [tex] \frac{dx}{dt} +3 \sqrt{x} \frac{dx}{dt} = 2t + 1[/tex]
[tex] \frac{dx}{dt} =\frac{2t + 1}{3 \sqrt{x}+1}[/tex]
dy/dt: [tex] \sqrt{t+1} \frac{dy}{dt} + \frac {y}{2\sqrt{t+1}} + \frac{t}{\sqrt{y}}\frac{dy}{dt}+ 2\sqrt{y} = 0[/tex]
just going to assign dummy variables:[tex] a = \sqrt{y}, b = \sqrt{t+1} [/tex] whilst I rearrange things...
[tex] \frac{dy}{dt}(b + \frac{t}{a}) = -(\frac {y}{2b} +2a)[/tex]
[tex] \frac{dy}{dt}(\frac{ba +t}{a}) = -(\frac {y +4ab}{2b})[/tex]
[tex] \frac{dy}{dt} = -(\frac {a(y +4ab)}{2b(ba+t)}) = -(\frac {\sqrt{y}(y + 4\sqrt{y}\sqrt{t+1})}{2\sqrt{t+1}(\sqrt{y}\sqrt{t+1}+t)})[/tex]
dy/dx =[tex] -\frac {(3 \sqrt{x}+1)(\sqrt{y}(y + 4\sqrt{y}\sqrt{t+1}))} {(2t + 1)(2\sqrt{t+1}(\sqrt{y}\sqrt{t+1}+t))}[/tex]
By using t = 0 and finding x and y in the original equations I get x = 1/4 and y = 4
[tex] -\frac{(\frac{3}{2}+1)(8 + 16)}{4} = -15 [/tex]
The answer I'm looking for however is -6. Is there some screw up with my working somewhere or am I using the wrong method?
I cannot graph any similar (but simpler) curves with Maxima to find the correct method neither because there doesn't seem to be a way to graph implicit expressions...nor will it differentiate one
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