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Question: If \(x=a{{t}^{2}},y=2at\), then find the value of \(\dfrac{dy}{dx}\)....

If x=at2,y=2atx=a{{t}^{2}},y=2at, then find the value of dydx\dfrac{dy}{dx}.

Explanation

Solution

Hint: First find the derivative of ‘x’ and ‘y’ with respect to ‘t’ and then use the formula dydx=dydtdxdt\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} to find the derivative of ‘y’ with respect to ‘x’.

Complete step-by-step answer:
To find the derivative of ‘y’ with respect to ‘x’ we will write the given equations first,
x=at2,y=2atx=a{{t}^{2}},y=2at

As ‘y’ and ‘x’ are defined in the form of an independent parameter ‘t’ therefore we have to use the method to find derivatives of parametric form.
For that we have to take the derivatives of ‘x’ and ‘y’ with respect to‘t’ so that we can get the dydx\dfrac{dy}{dx} by using a simple formula.

Therefore we will first find the derivative of ‘x’ with respect to ‘t’
x=at2x=a{{t}^{2}}

Differentiating above equation with respect to ‘t’ we will get,
dxdt=ddt(at2)\therefore \dfrac{dx}{dt}=\dfrac{d}{dt}\left( a{{t}^{2}} \right)
As ‘a’ is a constant therefore we can take it outside the derivative, therefore we will get,
dxdt=a×ddt(t2)\therefore \dfrac{dx}{dt}=a\times \dfrac{d}{dt}\left( {{t}^{2}} \right) ……………………………. (1)

To proceed further in the solution we should know the formula given below,
Formula:
ddx(xn)=n×xn1\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}
By using the formula given above we can write the equation (1) as,
dxdt=a×(2t21)\therefore \dfrac{dx}{dt}=a\times \left( 2{{t}^{2-1}} \right)
dxdt=a×(2t)\therefore \dfrac{dx}{dt}=a\times \left( 2t \right)
dxdt=2at\therefore \dfrac{dx}{dt}=2at ……………………………….. (2)

Now we will first find the derivative of ‘y’ with respect to ‘t’
y=2aty=2at
Differentiating above equation with respect to ‘t’ we will get,
dydt=ddt(2at)\therefore \dfrac{dy}{dt}=\dfrac{d}{dt}\left( 2at \right)
As ‘2a’ is a constant therefore we can take it outside the derivative, therefore we will get,
dydt=2a×ddt(t)\therefore \dfrac{dy}{dt}=2a\times \dfrac{d}{dt}\left( t \right) ……………………………. (3)

To proceed further in the solution we should know the formula given below,
Formula:
ddx(x)=1\dfrac{d}{dx}\left( x \right)=1
By using the formula given above we can write the equation (3) as,
dydt=2a×(1)\therefore \dfrac{dy}{dt}=2a\times \left( 1 \right)
dydt=2a\therefore \dfrac{dy}{dt}=2a……………………………….. (4)

Now, to find the derivative of ‘y’ with respect to ‘x’ we should know the formula given below,
Formula:
If ‘x’ and ‘y’ are functions of an independent parameter ‘t’ then, derivative of ‘y’ with respect to ‘x’ can be given as,
dydx=dydtdxdt\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}

If we put the values of equation (2) and equation (4) in above formula we will get,
dydx=2a2at\therefore \dfrac{dy}{dx}=\dfrac{2a}{2at}
By cancelling ‘2a’ from the numerator and denominator of the right hand side of the above equation we will get,
dydx=1t\therefore \dfrac{dy}{dx}=\dfrac{1}{t}
Therefore the value of dydx\dfrac{dy}{dx} is equal to 1t\dfrac{1}{t}.

Note: Don’t use the formula dydx=dydtdxdt\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} directly as it will complicate the solution. First calculate the values separately and then put them in formula for simplicity.