Question

A curve in the xy-plane is defined by the parametric equations `x = t^3 + 2` and `y = t^2 - 5t` how do you find the slope of the line tangent to the curve at the point where x = 10?

Answer 1

The slope of the tangent is `-1/12`

Explanation:

The curve is defined by the parametric equations:

`{ (x=t^3+2), (y=t^2-5t) :}`

When `x=10` then:

`x=t^3+2 => 10=t^3+2`
`:. t^3=8`
`:. \ \ t=2`

When `t=2` then:

`y=t^2-5t => y=4-10 = -6`

So the coordinate of interest with `t=2` is `(10,-6)`

We now need the derivative to establish the gradient of the tangent:

`x=t^3+2 \ => dx/dt=3t^2`
`y=t^2-5t => dy/dt=2t-5`

By the chain rule we have:

`dy/dx = (dy/dt)/(dx/dt)`

`" " = (2t-5)/(3t^2)`

So when `t=2` we have:

`dy/dx = -1/12`

So our tangent has slope `-1/12` and passes through `(10,-6)`, So using the point/slope formula `y-y_1=m(x-x_1)` for s straight line the required equation is:

`y-(-6)=-1/12(x-10)`
`:. y+6=-1/12x+10/12`
`:. y=-1/12x-31/6`

We can confirm this graphically: