# What is the slope of f(t) = (t+1,t^3) at t =1?

Jan 24, 2018

$f ' \left(x\right) = 3$

#### Explanation:

So we have $\left(t + 1 , {t}^{3}\right)$
This really means that:
$x = t + 1$ and $y = {t}^{3}$

We find $t$ in terms of $x$.
$x = t + 1$
$\implies t = x - 1$

We now plug this value in our function $y = {t}^{3}$.
$y = {\left(x - 1\right)}^{3}$ We simplify this.
$y = {\left(x - 1\right)}^{3} \implies y = {x}^{3} - 3 {x}^{2} + 3 x - 1$
We now find the derivative of this using the power rule.
$f ' \left(x\right) = 3 {x}^{2} - 6 x + 3$

Now, when $t = 1$, we see that:
$x = t + 1$
=>$x = 2$
We plug this into our derivative:
$f ' \left(x\right) = 3 {\left(2\right)}^{2} - 6 \left(2\right) + 3$
=>$f ' \left(x\right) = 12 - 12 + 3$
$f ' \left(x\right) = 12 - 12 + 3$
$f ' \left(x\right) = 3$ That is our answer!

Jan 24, 2018

$\text{slope } = 3$

#### Explanation:

•color(white)(x)m_(color(red)"tangent")=dy/dx" at t = 1"

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$x = t + 1 \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}} = 1 \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}} = 1$

$y = {t}^{3} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}} = 3 {t}^{2}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {t}^{2}$

$t = 1 \to \frac{\mathrm{dy}}{\mathrm{dx}} = 3$