# A particle moves along the line y = 5x - 2 in such a way that its x-coordinate at time t is x = 3t. Find dy/dt.

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We are given that y = 5x – 2. Also x = 3t, substituting x in the equation for y we get y = 5*3t – 2 or y = 15t – 2.

We can differentiate this with respect to t to get dy/dt

=d/dt (15t – 2)

= 15

This can also be done using the chain rule.

First we use y = 5x – 2 and differentiate dy/ dx, this gives d/dx (5x – 2) = 5.

Next we use x = 3t and differentiate dx/ dt which gives us d/dt (3t) = 3.

So dy/dt

= (dy/dx) (dx/dt)

= 5*3

= 15.

**Therefore using both the methods we get the result as 15.**

Given that a particle moves along the line:

y= 5x - 2

Also given that the change in x-coordinate to the time t is:

x = 3t

We need to determine dy/dt

We know that using the chain rule :

dy/dt = dy/dx * dx/ dt

Given y= 5x - 2

==> dy/dx = 5............(1)

Also, given x = 3t

==> dx/dt = 3 .........(2)

Then we will substitute (1) and (2) in the chain rule:

dy/dt = dy/dx * dx/ dt

= 5 * 3 = 15

**Then dy/dt = 15**

An alternative method of solving this problem is to substitute directly into y(x) to find y(t).

y(x) = 5x - 2

But we also know that x = 3t

Substituting, we have:

y(x(t)) = 5 * 3t - 2 = 15t - 2

Now, take the derivative to find dy/dt:

d/dt ( y ) = d/dt( 15t - 2 )

dy/dt = 15

This substitution method works in many cases, and in it you can see why it is important when using the chain rule to multiply by the derivative of the inside function.