Answer :
[tex]y = 24 + x^{3}[/tex]
[tex]\frac{dy}{dx} = 3x^{2} \text{, but } \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt} \text{ using the chain rule}[/tex]
[tex]\therefore 3x^{2} = \frac{dy}{dt} \cdot \frac{dx}{dt}[/tex]
Now, before we continue, I want to stress the importance of what dy/dt and dx/dt actually mean.
When we're dealing with time, space, and movement along a one-directional plane, we're moving with two scalar quantities (using our parametric representation of a particle), namely its x-direction and its y-direction.
Thus, by differentiating x with respect to t, we're literally saying: "well, what is the change of rate of x in relation to time", or "as t moves along the plane, what is the change of rate of x in that time?"
This can be stressed for the y-ordinate, as well.
So, the change of rate of the y-ordinate (dy/dt is 4cm/s). This is how fast the vertical quantity is changing as time changes at that point.
[tex]\therefore 3x^{2} = 4 \cdot \frac{dx}{dt}[/tex]
[tex]\frac{3x^{2}}{4} = \frac{dx}{dt}[/tex]
[tex]\therefore \frac{dx}{dt} = \frac{3(1)}{4} = \frac{3}{4} \text{ at } x = 1[/tex]
[tex]\frac{dy}{dx} = 3x^{2} \text{, but } \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dx}{dt} \text{ using the chain rule}[/tex]
[tex]\therefore 3x^{2} = \frac{dy}{dt} \cdot \frac{dx}{dt}[/tex]
Now, before we continue, I want to stress the importance of what dy/dt and dx/dt actually mean.
When we're dealing with time, space, and movement along a one-directional plane, we're moving with two scalar quantities (using our parametric representation of a particle), namely its x-direction and its y-direction.
Thus, by differentiating x with respect to t, we're literally saying: "well, what is the change of rate of x in relation to time", or "as t moves along the plane, what is the change of rate of x in that time?"
This can be stressed for the y-ordinate, as well.
So, the change of rate of the y-ordinate (dy/dt is 4cm/s). This is how fast the vertical quantity is changing as time changes at that point.
[tex]\therefore 3x^{2} = 4 \cdot \frac{dx}{dt}[/tex]
[tex]\frac{3x^{2}}{4} = \frac{dx}{dt}[/tex]
[tex]\therefore \frac{dx}{dt} = \frac{3(1)}{4} = \frac{3}{4} \text{ at } x = 1[/tex]