Answer
Verified
396.6k+ views
Hint: In order to solve this problem, we need to know the chain rule. The chain rule is given by $\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$ . Also, in order to simplify the equation, we need to know some formulas. They are given by $\sin 2x=2\sin x.\cos x$, $1-\cos 2x=2{{\sin }^{2}}x$ and $\cot x=\dfrac{\cos x}{\sin x}$ .
Complete step by step answer:
As we can see that the $y$ is the function of $t$ . and $x$ is also the function of $t$ .
Hence, we cannot find the value $\dfrac{dy}{dx}$ directly by differentiating $y$ .
To solve this, we need to use the chain rule.
The chain rule says that
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}...........(i)$
Therefore, we now need to find the values of $\dfrac{dy}{dt}$ and $\dfrac{dt}{dx}$ separately.
Differentiating $y=12\left( 1-\cos t \right)$ we get,
$\dfrac{dy}{dt}=\dfrac{d}{dt}\left( 12\left( 1-\cos t \right) \right)$
Solving this further we get,
\[\begin{align}
& \dfrac{dy}{dt}=\dfrac{d}{dt}\left( 12-12\cos t \right) \\
& =\left( 0-12\left( -\sin t \right) \right) \\
& =12\sin t..........................(ii)
\end{align}\]
Similarly differentiating $x=10\left( t-\sin t \right)$ , we get,
$\begin{align}
& \dfrac{dx}{dt}=\dfrac{d}{dt}\left( 10\left( t-\sin t \right) \right) \\
& \\
\end{align}$
Solving this further we get,
$\begin{align}
& \dfrac{dx}{dt}=\dfrac{d}{dt}\left( 10\left( t-\sin t \right) \right) \\
& =\dfrac{d}{dt}\left( 10t-10\sin t \right) \\
& =10-10\cos t.....................(iii)
\end{align}$
We need to find the value of $\dfrac{dt}{dx}$ , taking the inverse of equation (iii) we get,
$\dfrac{dt}{dx}=\dfrac{1}{10-10\cos t}............(iv)$
Substituting the values of equation (ii) and (iv) in equation (i) we get,
$\dfrac{dy}{dx}=12\sin t\times \dfrac{1}{10-10\cos t}$
Solving this we get,
$\begin{align}
& \dfrac{dy}{dx}=\dfrac{12\sin t}{10\left( 1-\cos t \right)} \\
& =\dfrac{6\sin t}{5\left( 1-\cos t \right)}
\end{align}$
We can use the formulas $\sin 2x=2\sin x.\cos x$ and $1-\cos 2x=2{{\sin }^{2}}x$ .
Using the formula, we get,
$\dfrac{dy}{dx}=\dfrac{6\left( 2\sin \dfrac{t}{2}.\cos \dfrac{t}{2} \right)}{5\left( 2{{\sin }^{2}}\dfrac{t}{2} \right)}$
Solving this further we get,
$\begin{align}
& \dfrac{dy}{dx}=\dfrac{6\left( \cos \dfrac{t}{2} \right)}{5\left( \sin \dfrac{t}{2} \right)} \\
& =\dfrac{6}{5}\cot \dfrac{t}{2}
\end{align}$
Hence, the value of $\dfrac{dy}{dx}=\dfrac{6}{5}\cot \dfrac{t}{2}$.
Note: We can also solve this dividing equation (ii) by (iii). The answer will remain the same. We can also solve this by another approach. We can solve this by finding the value of $t$ in terms of $x$ and substituting in the equation of $y$ . Our aim will be to eliminate the value of $t$ . But we need to be careful as this method can turn out to be extremely complicated.
Complete step by step answer:
As we can see that the $y$ is the function of $t$ . and $x$ is also the function of $t$ .
Hence, we cannot find the value $\dfrac{dy}{dx}$ directly by differentiating $y$ .
To solve this, we need to use the chain rule.
The chain rule says that
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}...........(i)$
Therefore, we now need to find the values of $\dfrac{dy}{dt}$ and $\dfrac{dt}{dx}$ separately.
Differentiating $y=12\left( 1-\cos t \right)$ we get,
$\dfrac{dy}{dt}=\dfrac{d}{dt}\left( 12\left( 1-\cos t \right) \right)$
Solving this further we get,
\[\begin{align}
& \dfrac{dy}{dt}=\dfrac{d}{dt}\left( 12-12\cos t \right) \\
& =\left( 0-12\left( -\sin t \right) \right) \\
& =12\sin t..........................(ii)
\end{align}\]
Similarly differentiating $x=10\left( t-\sin t \right)$ , we get,
$\begin{align}
& \dfrac{dx}{dt}=\dfrac{d}{dt}\left( 10\left( t-\sin t \right) \right) \\
& \\
\end{align}$
Solving this further we get,
$\begin{align}
& \dfrac{dx}{dt}=\dfrac{d}{dt}\left( 10\left( t-\sin t \right) \right) \\
& =\dfrac{d}{dt}\left( 10t-10\sin t \right) \\
& =10-10\cos t.....................(iii)
\end{align}$
We need to find the value of $\dfrac{dt}{dx}$ , taking the inverse of equation (iii) we get,
$\dfrac{dt}{dx}=\dfrac{1}{10-10\cos t}............(iv)$
Substituting the values of equation (ii) and (iv) in equation (i) we get,
$\dfrac{dy}{dx}=12\sin t\times \dfrac{1}{10-10\cos t}$
Solving this we get,
$\begin{align}
& \dfrac{dy}{dx}=\dfrac{12\sin t}{10\left( 1-\cos t \right)} \\
& =\dfrac{6\sin t}{5\left( 1-\cos t \right)}
\end{align}$
We can use the formulas $\sin 2x=2\sin x.\cos x$ and $1-\cos 2x=2{{\sin }^{2}}x$ .
Using the formula, we get,
$\dfrac{dy}{dx}=\dfrac{6\left( 2\sin \dfrac{t}{2}.\cos \dfrac{t}{2} \right)}{5\left( 2{{\sin }^{2}}\dfrac{t}{2} \right)}$
Solving this further we get,
$\begin{align}
& \dfrac{dy}{dx}=\dfrac{6\left( \cos \dfrac{t}{2} \right)}{5\left( \sin \dfrac{t}{2} \right)} \\
& =\dfrac{6}{5}\cot \dfrac{t}{2}
\end{align}$
Hence, the value of $\dfrac{dy}{dx}=\dfrac{6}{5}\cot \dfrac{t}{2}$.
Note: We can also solve this dividing equation (ii) by (iii). The answer will remain the same. We can also solve this by another approach. We can solve this by finding the value of $t$ in terms of $x$ and substituting in the equation of $y$ . Our aim will be to eliminate the value of $t$ . But we need to be careful as this method can turn out to be extremely complicated.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The cell wall of prokaryotes are made up of a Cellulose class 9 biology CBSE
What organs are located on the left side of your body class 11 biology CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE