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Question: How do you find the equations of the tangent and normal of the curve at \(x={{t}^{2}}\), \(y=t+3\), ...

How do you find the equations of the tangent and normal of the curve at x=t2x={{t}^{2}}, y=t+3y=t+3, t=1t=1?

Explanation

Solution

In this question we have to find the slope and points of the tangent and the normal of the curve. We will first take the derivative of the term x=t2x={{t}^{2}} with respect to tt. Then we will take the derivative of y=t+3y=t+3 with respect to tt. Then divide both to get dydx\dfrac{dy}{dx} which is the slope and substitute the value of t=1t=1 to get the value of the slope. We will then use point slope form to find the equation of tangent and then use the formula mn=−1m{{m}_{n}}=\dfrac{-1}{m} to find the slope of the normal and use point slope form to get its equation.

Complete step by step solution:
We have the curve as x=t2x={{t}^{2}} and y=t+3y=t+3.
On differentiating the term x=t2x={{t}^{2}}, we get:
⇒dxdt=ddt(t2)\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{t}^{2}} \right)
On using the formula ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}, we get:
⇒dxdt=2t\Rightarrow \dfrac{dx}{dt}=2t
On differentiating the term y=t+3y=t+3, we get:
⇒dydt=ddt(t+3)\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( t+3 \right)
On splitting the derivative, we get:
⇒dydx=dtdt+ddt3\Rightarrow \dfrac{dy}{dx}=\dfrac{dt}{dt}+\dfrac{d}{dt}3
On using the formula dxdx=1\dfrac{dx}{dx}=1 and ddxk=0\dfrac{d}{dx}k=0, we get:
⇒dydt=1\Rightarrow \dfrac{dy}{dt}=1
We know that tangent of a curve is given by dydx\dfrac{dy}{dx} therefore, we can write:
⇒dydtdxdt=12t\Rightarrow \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{1}{2t}
On simplifying, we get:
⇒dydx=12t\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2t}
Now at t=1t=1, we have:
⇒dydx∣t=1=12(1)\Rightarrow {{\left. \dfrac{dy}{dx} \right|}_{t=1}}=\dfrac{1}{2\left( 1 \right)}
On simplifying, we get:
⇒dydx=12\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}
Which is the required slope therefore, m=12m=\dfrac{1}{2}.
Now at t=1t=1, xx will be:
⇒x=(1)2=1\Rightarrow x={{\left( 1 \right)}^{2}}=1
And at t=1t=1, yy will be:
⇒y=1+3=4\Rightarrow y=1+3=4
So, the tangent passes through (1,4)\left( 1,4 \right) and has slope 12\dfrac{1}{2}, so using the point slope form which is y−y1=m(x−x1)y-{{y}_{1}}=m\left( x-{{x}_{1}} \right), we get:
⇒y−4=12(x−1)\Rightarrow y-4=\dfrac{1}{2}\left( x-1 \right)
On simplifying, we get:
⇒y=12x+72\Rightarrow y=\dfrac{1}{2}x+\dfrac{7}{2}
And the normal passes through (1,4)\left( 1,4 \right) and has slope mn=−1m=−11/2=−2{{m}_{n}}=\dfrac{-1}{m}=\dfrac{-1}{1/2}=-2, so using the point slope form, we get:
⇒y−4=−2(x−1)\Rightarrow y-4=-2\left( x-1 \right)
On simplifying, we get:
⇒y=−2x+6\Rightarrow y=-2x+6
On drawing the curve, the tangent and normal on the graph, we get:

Note: This type of question belongs to the category of calculus. the addition rule of differentiation should be remembered which is dydx(f(x)+g(x))=ddx(f(x))+ddx(g(x))\dfrac{dy}{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). It is to be remembered that the derivative of the equation of the line gives the slope of the line and a normal is a line perpendicular to the tangent of a curve.