What is the slope of the tangent line of #(y/x)e^(x/y-x^2)= C #, where C is an arbitrary constant, at #(1,2)#?

Question

Isaiah Allen · Accepted Answer

10

Taking the logarithm of both sides gives us #ln y -ln x +x/y-x^2 = ln C# Differentiating with respect to #x# yields #1/y dy/dx -1/x +1/y-x/y^2dy/dx-2x = 0 implies# #(1/y-x/y^2)dy/dx =2x+1/x-1/y implies # #x(y-x)dy/dx = (2x^2+1)y^2-xy# At the point #(1,2)# we have #1 times (2-1)times dy/dx = (2 times1^2+1)times 2^2-1times 2 implies# # dy/dx = 10# Hence the slope of the tangent at #(1,2)# is 10 Note I have assumed that tangent at #(1,2)# means tangent to the curve at #(1,2)#. Of course, this means that the constant #C# can not be arbitrary - but rather has to be #2/sqrt e#

Christopher Agresta · Answer

undefined To find the slope of the tangent line at the point $(1,2)$ for the curve $ \frac{y}{x} e^{(x/y - x^2)} = C $, differentiate the equation implicitly with respect to $x$ and $y$.

Differentiating implicitly with respect to $x$:
$$ \frac{y}{x} e^{(x/y - x^2)} + \frac{dy}{dx} e^{(x/y - x^2)} \left(1 - \frac{2x}{y}\right) = 0 $$

Differentiating implicitly with respect to $y$:
$$ \frac{1}{x} e^{(x/y - x^2)} + \frac{y}{x^2} e^{(x/y - x^2)} \left(1 - \frac{x^2}{y^2}\right) \frac{dx}{dy} = 0 $$

Evaluate both equations at the point $(1,2)$ to find $\frac{dy}{dx}$ and $\frac{dx}{dy}$.

Using these values, the slope $m$ of the tangent line is given by:
$$ m = -\frac{dx}{dy} \times \frac{y}{\frac{dy}{dx}} $$

After calculating, the slope $m$ at $(1,2)$ is:

$$ m = -\frac{dx}{dy} \times \frac{2}{\frac{dy}{dx}} $$

Solving for $m$:

$$ m = -\frac{e^{(1/2 - 1)}}{1 + \frac{4}{2^2}} \times \frac{2}{e^{(1/2 - 1)} \left(1 - \frac{2(1)}{2}\right)} $$

$$ m = -\frac{e^{-1/2}}{1 + 1} \times \frac{2}{e^{-1/2} \left(1 - 1\right)} $$

$$ m = -\frac{e^{-1/2}}{2} \times \frac{2}{0} $$

Since the denominator is 0, the slope is undefined at the point $(1,2)$.

Luke Alvarez · Answer

undefined To find the slope of the tangent line of the curve $y/x e^{x/y - x^2} = C$ at the point (1,2), we first need to find the equation of the curve and then differentiate it implicitly with respect to $x$.

Given: $y/x e^{x/y - x^2} = C$

Taking the natural logarithm of both sides, we get:

$$\ln\left(\frac{y}{x} e^{x/y - x^2}\right) = \ln(C)$$

Using logarithmic properties, we simplify to:

$$\ln(y) - \ln(x) + \frac{x}{y} - x^2 = \ln(C)$$

Now, differentiating implicitly with respect to $x$, we get:

$$\frac{1}{y} \frac{dy}{dx} - \frac{1}{x} - \frac{x}{y^2} \frac{dy}{dx} - 2x = 0$$

At the point (1,2), we substitute $x = 1$ and $y = 2$ into the equation above to find the slope of the tangent line. After substituting, we solve for $\frac{dy}{dx}$, which gives us the slope of the tangent line at that point.