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Question: If \({{x}^{2}}+{{y}^{2}}=25\) and \(\dfrac{dy}{dt}=6\), how do you find \(\dfrac{dx}{dt}\) when \(y=...

If x2+y2=25{{x}^{2}}+{{y}^{2}}=25 and dydt=6\dfrac{dy}{dt}=6, how do you find dxdt\dfrac{dx}{dt} when y=4y=4 ?

Explanation

Solution

In order to find a solution to this problem, we will use implicit differentiation. For implicit differentiation, the process will go like differentiating all terms with respect to timetime, use derivative rules and then solve for dxdt\dfrac{dx}{dt}.

Complete step by step solution:
From the above problem as we can see that the function cannot be solved explicitly, so we will use implicit differentiation.
Implicit differentiation is used when the function cannot be solved explicitly.
We have our expression as:
x2+y2=25{{x}^{2}}+{{y}^{2}}=25
By differentiating our expression with respect to time, we get:
ddt(x2+y2)=ddt(25)\dfrac{d}{dt}\left( {{x}^{2}}+{{y}^{2}} \right)=\dfrac{d}{dt}\left( 25 \right)
2xdxdt+2ydydt=02x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0
Now by dividing 2x2x on both side, we get:
2x×12x×dxdt+2y×12x×dydt=0×12x2x\times \dfrac{1}{2x}\times \dfrac{dx}{dt}+2y\times \dfrac{1}{2x}\times \dfrac{dy}{dt}=0\times \dfrac{1}{2x}
On Simplifying, we get:
dxdt+yxdydt=0\dfrac{dx}{dt}+\dfrac{y}{x}\dfrac{dy}{dt}=0
Now by subtracting yx\dfrac{y}{x} on both the side, we get:
dxdt=yxdydt(1)\Rightarrow \dfrac{dx}{dt}=-\dfrac{y}{x}\dfrac{dy}{dt}\to \left( 1 \right)
Now, substituting dydt=6\dfrac{dy}{dt}=6 in equation (1)\left( 1 \right),
dxdt=6y25y2(2)\Rightarrow \dfrac{dx}{dt}=-6\dfrac{y}{\sqrt{25-{{y}^{2}}}}\to \left( 2 \right)
Therefore, we get the rate of change in xx as a function of xx and yy. Butxx and yyare related by the given equation: x2+y2=25{{x}^{2}}+{{y}^{2}}=25
Eliminating xx and thus expressing dxdt\dfrac{dx}{dt} as a function of yy only.
Now as we have all the terms, we now have to substitute in equation (2).
That is, we have y=4y=4
dxdt=6y25y2\Rightarrow \dfrac{dx}{dt}=-6\dfrac{y}{\sqrt{25-{{y}^{2}}}}
Now on substituting y=4y=4we get:
dxdt=642542\Rightarrow \dfrac{dx}{dt}=-6\dfrac{4}{\sqrt{25-{{4}^{2}}}}
On simplifying,
dxdt=642516\Rightarrow \dfrac{dx}{dt}=-6\dfrac{4}{\sqrt{25-16}}
dxdt=649\Rightarrow \dfrac{dx}{dt}=-6\dfrac{4}{\sqrt{9}}
On taking root of 9=3\sqrt{9}=3, we have:
dxdt=643\Rightarrow \dfrac{dx}{dt}=-6\dfrac{4}{3}
Therefore, on further simplification, we get:
dxdt=8\Rightarrow \dfrac{dx}{dt}=-8
Therefore, dxdt\dfrac{dx}{dt} when y=4y=4in expression x2+y2=25{{x}^{2}}+{{y}^{2}}=25 is:
dxdt=8\Rightarrow \dfrac{dx}{dt}=-8
Therefore,
dxdt=8\Rightarrow \dfrac{dx}{dt}=-8 is the required solution.

Note:
Implicit differentiation is useful when it is difficult to manipulate the equation such that yyis isolated on one side of the equation.
To differentiate implicitly, we have to differentiate both the sides of the equation. Then, we have to use relevant derivative rules on all terms and then solve for dydx\dfrac{dy}{dx}.
Also, while doing derivatives we have to be sure whether from which term we have to derivate. We got confused about which term we have to derivate, so we have to be careful about that.