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Question: The equation of the curve passing through \(\left( {1,3} \right)\) whose slope at any point \(\left(...

The equation of the curve passing through (1,3)\left( {1,3} \right) whose slope at any point (x,y)\left( {x,y} \right) on it is yx2\dfrac{y}{{{x^2}}} is given by:
A. y=3e−1xy = 3{e^{ - \dfrac{1}{x}}}
B. y=3e1−1xy = 3{e^{1 - \dfrac{1}{x}}}
C. y=ce1xy = c{e^{\dfrac{1}{x}}}
D. y=3e1xy = 3{e^{\dfrac{1}{x}}}

Explanation

Solution

Here we must know that slope is given by the differentiation of the function given and hence we can say that in this one function we are given dydx=yx2\dfrac{{dy}}{{dx}} = \dfrac{y}{{{x^2}}} and now we can integrate this to get the required function and the constant of integration can be calculating by substituting the point (1,3)\left( {1,3} \right)

Complete step by step answer:
Here we are given that the equation of the curve passing through (1,3)\left( {1,3} \right) whose slope at any point (x,y)\left( {x,y} \right) on it is yx2\dfrac{y}{{{x^2}}}
We must know that slope is given by the differentiation of the function given and hence we can say that in this one function we are given
dydx=yx2\dfrac{{dy}}{{dx}} = \dfrac{y}{{{x^2}}}
Integrating both the sides we get that:
∫dydx=∫yx2\int {\dfrac{{dy}}{{dx}} = \int {\dfrac{y}{{{x^2}}}} }
∫dyy=∫dxx2\int {\dfrac{{dy}}{y} = \int {\dfrac{{dx}}{{{x^2}}}} }
log⁡y=−1x+c y=e−1x+c  \log y = - \dfrac{1}{x} + c \\\ y = {e^{ - \dfrac{1}{x} + c}} \\\
Now we can write the above as:
y=e−1xecy = {e^{ - \dfrac{1}{x}}}{e^c}
Now we can take ec{e^c} as another constant which can be let as kk
So we get:
y=e−1xky = {e^{ - \dfrac{1}{x}}}k
As we know that the curve passes through the point (1,3)\left( {1,3} \right) as it is given in the problem. Therefore we can substitute this point in eth above problem and we will get the solution as:
3=e−11k 3=ke k=3e  3 = {e^{ - \dfrac{1}{1}}}k \\\ 3 = \dfrac{k}{e} \\\ k = 3e \\\
Now substituting this value in the above function where we have kk we get:
y=e−1x(3e) y=3e1−1x  y = {e^{ - \dfrac{1}{x}}}\left( {3e} \right) \\\ y = 3{e^{1 - \dfrac{1}{x}}} \\\
We can add the powers of the same base if they are being multiplied like (am)(an)=am+n\left( {{a^m}} \right)\left( {{a^n}} \right) = {a^{m + n}}
So we get the function as y=3e1−1xy = 3{e^{1 - \dfrac{1}{x}}}

So, the correct answer is “Option B”.

Note: Here the student must know that whenever we are given slope in the form of the variable then we can easily find the slope by finding the derivative of the function and if we are given the angle of inclination of the line which the positive x−axisx - {\text{axis}} then slope is given by tangent of the angle of inclination.