Question

Find the slope of the tangent to the curve $x=a{{t}^{2}}$ and $y=2at$ at t=1.

Answer 1

We solve this problem by first discussing the property that is slope of tangent at point $\left( {{x}_{1}},{{y}_{1}} \right)$ on the curve $y=f\left( x \right)$ is equal to ${{\left. \dfrac{dy}{dx} \right]}_{\left( {{x}_{1}},{{y}_{1}} \right)}}$. Then we find the values of $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ by differentiating $y=2at$ and $x=a{{t}^{2}}$ with respective to t using the formula, $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$. Then we substitute this values in $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$ and then substitute the value $t=1$ in it to find the slope of the required tangent.

Complete step-by-step solution:
The curve we are given is $x=a{{t}^{2}}$ and $y=2at$.
We need to find the slope of the tangent to this curve at t=1.
Here the equation of the curve is given in the parametric form.
Now let us remember the property that,
The slope of any curve y given by $y=f\left( x \right)$ at any point $\left( {{x}_{1}},{{y}_{1}} \right)$ is equal to the value of $\dfrac{dy}{dx}$ at that point, that is slope of tangent at $\left( {{x}_{1}},{{y}_{1}} \right)$ on the curve is equal to ${{\left. \dfrac{dy}{dx} \right]}_{\left( {{x}_{1}},{{y}_{1}} \right)}}$.

Here the curve we are given is $x=a{{t}^{2}}$ and $y=2at$.
We need to find the slope of the tangent to this curve at point t=1. From the above-discussed property,
Slope of the tangent = ${{\left. \dfrac{dy}{dx} \right]}_{t=1}}$
So first, let us find the value of $\dfrac{dy}{dx}$ for the given curve.
We can write $\dfrac{dy}{dx}$ as,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}............\left( 1 \right)$
Now let us consider $y=2at$.
Let us differentiate it with respect to t.
Let us consider the formula for differentiation,
$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
Using this formula, we get,
$\Rightarrow \dfrac{dy}{dt}=2a.........\left( 2 \right)$
Now let us consider $x=a{{t}^{2}}$.
Let us differentiate it with respect to t.
Let us consider the formula for differentiation,
$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
Using this formula, we get,
$\begin{aligned} & \Rightarrow \dfrac{dx}{dt}=a\left( 2t \right) \\\ & \Rightarrow \dfrac{dx}{dt}=2at.........\left( 3 \right) \\\ \end{aligned}$
Substituting the values in equations (2) and (3) in equation (1) we get,
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{2at} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{t} \\\ \end{aligned}$
As we need to find the value of ${{\left. \dfrac{dy}{dx} \right]}_{t=1}}$ let us substitute the value t=1 in the above obtained value of $\dfrac{dy}{dx}$. Then we get,
$\begin{aligned} & \Rightarrow {{\left. \dfrac{dy}{dx} \right]}_{t=1}}={{\left. \dfrac{1}{t} \right]}_{t=1}} \\\ & \Rightarrow {{\left. \dfrac{dy}{dx} \right]}_{t=1}}=1 \\\ \end{aligned}$
So, we get that slope of the tangent to given curve at t=1 is equal to 1.
Hence answer is 1.

Note: We can also solve this question in another process.
We need to find the slope of the tangent to the curve $x=a{{t}^{2}}$ and $y=2at$ at point t=1.
Now let us find the equation of curve
$\begin{aligned} & \Rightarrow x=a{{t}^{2}} \\\ & \Rightarrow {{t}^{2}}=\dfrac{x}{a} \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow y=2at \\\ & \Rightarrow t=\dfrac{y}{2a} \\\ \end{aligned}$
Using these two values we get,
$\begin{aligned} & \Rightarrow {{\left( \dfrac{y}{2a} \right)}^{2}}=\dfrac{x}{a} \\\ & \Rightarrow \dfrac{{{y}^{2}}}{4{{a}^{2}}}=\dfrac{x}{a} \\\ & \Rightarrow {{y}^{2}}=4ax \\\ \end{aligned}$
At, t=1
$\begin{aligned} & \Rightarrow x=a{{t}^{2}}=a{{\left( 1 \right)}^{2}}=a \\\ & \Rightarrow y=2at=2a\left( 1 \right)=2a \\\ \end{aligned}$
So, we need to find the slope of tangent at the point $\left( a,2a \right)$.
First let us find the value of $\dfrac{dy}{dx}$.
Now let us consider the curve ${{y}^{2}}=4ax$.
Now let us differentiate with respective to x. Then we get,
Let us consider the formula for differentiation,
$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
Using this formula, we get,
$\begin{aligned} & \Rightarrow 2y\dfrac{dy}{dx}=4a \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{y} \\\ \end{aligned}$
Now let us find the value of ${{\left. \dfrac{dy}{dx} \right]}_{\left( a,2a \right)}}$.
$\Rightarrow {{\left. \dfrac{dy}{dx} \right]}_{\left( a,2a \right)}}=\dfrac{2a}{2a}=1$
So, we get the slope of a tangent as 1.
Hence answer is 1.