equation of a tangent line

[janeann]

Find the equation of the tangent line to f(t)=(tsin(t)+t^2cos(t))(2/1+t^3) at point (0,0).
For my work I have the derivative of the equation being:2(t^5+2t^3+t^3sin(t)-3cos(t))/(t^3-1)^2. After i found the derivatie i plugged in the point 0 in for t and i got 0 as my anwser and the equation for the tangent line being y=0. by using the y-0=0(m-0). But i have a feeling that this is not correct! please help.

 

[renegade05]

Find the equation of the tangent line to f(t)=(tsin(t)+t^2cos(t))(2/1+t^3) at point (0,0).
For my work I have the derivative of the equation being:2(t^5+2t^3+t^3sin(t)-3cos(t))/(t^3-1)^2. After i found the derivatie i plugged in the point 0 in for t and i got 0 as my anwser and the equation for the tangent line being y=0. by using the y-0=0(m-0). But i have a feeling that this is not correct! please help.

First, why do you have a feeling that is not correct?

Second, that is a disgusting equation to derive analytically.

Third, here is the graph of f(t). As you can see at the point (0,0) f'(t)=0. (Local Min)

So yes the tangent line equation is y=0

[attachment=0:g302jsah]photo.jpg[/attachment:g302jsah]

 

[janeann]

I thought that the way i plugged in 0 into the equation was wrong. But i guess not! thank you!!

 

[renegade05]

Find the equation of the tangent line to f(t)=(tsin(t)+t^2cos(t))(2/1+t^3) at point (0,0).
For my work I have the derivative of the equation being:2(t^5+2t^3+t^3sin(t)-3cos(t))/(t^3-1)^2

After closer inspection I realized your derivative is slightly incorrect.

should be :

\(\displaystyle f'(t)=-\frac{2((t^5+2t^3+t^2-1)\sin(t)-3t\cos(t))}{(t^3+1)^2}\)

Nevertheless the tangent line at (0,0) remains y=0

 

[janeann]

Thank you!!