Area pf a rectangle with s moving vertex

[Slithus]

A rectangle has 2 sides along the coordinate axes and one vertex at a point P which moves along the curve y = e^x in such a way that x increases at the rate of 1 cm / 2 sec. How fast is the area of the rectangle increasing when x = ln2 cm?

 

[topsquark]

[imath]A = xy = x e^x[/imath], so [imath]\dfrac{dA}{dt} = \dfrac{d}{dt} \left ( x e^x \right )[/imath] and you know dx/dt = (1/2) cm/s.

Can you take the derivative?

-Dan

 

[Slithus]

[imath]A = xy = x e^x[/imath], so [imath]\dfrac{dA}{dt} = \dfrac{d}{dt} \left ( x e^x \right )[/imath] and you know dx/dt = (1/2) cm/s.

Can you take the derivative?

-Dan

Is the answer 0.24 cm/s? I'm not sure I took the correct derivative. Other than that I keep getting 0.

 

[Deleted member 4993]

Is the answer 0.24 cm/s? I'm not sure I took the correct derivative. Other than that I keep getting 0.

Please share your work for getting the answer as 0.24 cm/s.

 

[Slithus]

Please share your work for getting the answer as 0.24 cm/s.

dx/dt xe^x = -((x^2)(e^x))/(t^2), -((ln2)^2)(e^(ln2))/4 = ∣-0.24∣

 

[topsquark]

dx/dt xe^x = -((x^2)(e^x))/(t^2), -((ln2)^2)(e^(ln2))/4 = ∣-0.24∣

You are taking a derivative. Use the product rule: [imath]\dfrac{d}{dt}(fg) = \dfrac{df}{dt} g + f \dfrac{dg}{dt}[/imath]

-Dan

 

[topsquark]

dx/dt xe^x = -((x^2)(e^x))/(t^2), -((ln2)^2)(e^(ln2))/4 = ∣-0.24∣

I should also mention that you are setting t = 2 s here. No one said that t = 2 s. dx/dt = (1/2) cm/s and we want dA/dt when x = ln(2).

-Dan