Question

How do you find the equation of the tangent to the curve $x={{t}^{4}}+1$, $y={{t}^{3}}+t$ at the point where $t=-1$?

Answer 1

This question is from the topic of calculus. In solving this question, we will differentiate the equation $x={{t}^{4}}+1$ with respect to t. After that, we will differentiate the equation $y={{t}^{3}}+t$ with respect to t. After that, we will find out the value of $\dfrac{dy}{dx}$. In the equation of $\dfrac{dy}{dx}$, we will put the value of t as -1. After that, we will use the formula of straight lines and get the answer.

Complete step-by-step solution:
Let us solve this question.
In this question, we have asked to find the equation of tangent of the curve. The curve is given as $x={{t}^{4}}+1$ and $y={{t}^{3}}+t$. We have to find the tangent at t=-1.
Let us first differentiate the equation $x={{t}^{4}}+1$ with respect to t, we can write
$\dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{t}^{4}}+1 \right)$
Using the addition rule of differentiation: $\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)\dfrac{d}{dx}\left( g\left( x \right) \right)$ [where f(x) and g(x) are two function of x], we can write
$\dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{t}^{4}} \right)+\dfrac{d}{dt}\left( 1 \right)$
As we know that differentiation of any constant is always zero, so we can write
$\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{t}^{4}} \right)+0$
Using the formula of differentiation: $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, we can write
$\Rightarrow \dfrac{dx}{dt}=4{{t}^{4-1}}$
$\Rightarrow \dfrac{dx}{dt}=4{{t}^{3}}$
Now, we will differentiate the equation $y={{t}^{3}}+t$ with respect to t, we can write
$\dfrac{dy}{dt}=\dfrac{d}{dt}\left( {{t}^{3}}+t \right)$
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( {{t}^{3}}+{{t}^{1}} \right)$
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( {{t}^{3}} \right)+\dfrac{d}{dt}\left( {{t}^{1}} \right)$
Using the formula: $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, we can write
$\Rightarrow \dfrac{dy}{dt}=3{{t}^{3-1}}+1$
$\Rightarrow \dfrac{dy}{dt}=3{{t}^{2}}+1$
As we know that the tangent of any curve is always equal to $\dfrac{dy}{dx}$, so we will divide the equation $\dfrac{dy}{dt}=3{{t}^{2}}+1$ by the equation $\dfrac{dx}{dt}=4{{t}^{3}}$, we can write
$\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3{{t}^{2}}+1}{4{{t}^{3}}}$
The above equation can also be written as
$\Rightarrow \dfrac{dy}{dt}\times \dfrac{dt}{dx}=\dfrac{3{{t}^{2}}+1}{4{{t}^{3}}}$
The above equation can also be written as
$\Rightarrow \dfrac{dy}{dx}=\dfrac{3{{t}^{2}}+1}{4{{t}^{3}}}$
Now, we will put here the value of t as -1, we will get
$\Rightarrow {{\left. \dfrac{dy}{dx} \right|}_{t=-1}}=\dfrac{3{{\left( -1 \right)}^{2}}+1}{4{{\left( -1 \right)}^{3}}}$
The above equation can also be written as
$\Rightarrow {{\left. \dfrac{dy}{dx} \right|}_{t=-1}}=\dfrac{3+1}{4\left( -1 \right)}=\dfrac{4}{-4}$
The above equation can also be written as
$\Rightarrow {{\left. \dfrac{dy}{dx} \right|}_{t=-1}}=-1$
Hence, we get the slope at t=-1 as -1.
Now, we will find the value of x and y at t=-1.
So, at t=-1, the value of x will be
$x={{\left( -1 \right)}^{4}}+1=1+1=2$
And, at t=-1, the value of y will be
$y={{\left( -1 \right)}^{3}}+\left( -1 \right)=-1-1=-2$
Now, using the co-ordinate $\left( {{x}_{1}},{{y}_{1}} \right)$ as (2,-2) and slope as -1, we can write the straight line as
$\left( y-{{y}_{1}} \right)=\left( \dfrac{dy}{dx} \right)\left( x-{{x}_{1}} \right)$
$\Rightarrow \left( y-\left( -2 \right) \right)=\left( -1 \right)\left( x-\left( 2 \right) \right)$
The above can also be written as
$\Rightarrow y+2=\left( -1 \right)\left( x-2 \right)$
$\Rightarrow y+2=-x+2$
$\Rightarrow x+y+2-2=0$
The above equation can also be written as
$\Rightarrow x+y=0$
Hence, the equation of tangent at t=-1 is $x+y=0$.

Note: We should have a better knowledge in the topic of calculus to solve this type of question easily. We should remember the following formulas of differentiation:
Addition rule of differentiation: $\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)\dfrac{d}{dx}\left( g\left( x \right) \right)$
$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
We should know that if a point is given as $\left( {{x}_{1}},{{y}_{1}} \right)$ and the slope is $\dfrac{dy}{dx}$, then the equation of line will be
$\left( y-{{y}_{1}} \right)=\left( \dfrac{dy}{dx} \right)\left( x-{{x}_{1}} \right)$