Challenge problem Find k if x=k is tangent to the curve y=x+√(2).e^[(x+y)/√(2)]

In summary, the conversation is discussing the value of k if the line x=k is tangent to the curve y=x+√2e^((x+y)/√2). An alternative solution is mentioned that does not require calculus, but it is left for others to try. The value of k is the main question of the conversation.
  • #1
Olinguito
239
0
If the line $x=k$ is tangent to the curve
$$\large y\:=\:x+\sqrt2\,e^{\frac{x+y}{\sqrt2}}$$
what is the value of $k$?
 
Mathematics news on Phys.org
  • #2
Olinguito said:
If the line $x=k$ is tangent to the curve
$$\large y\:=\:x+\sqrt2\,e^{\frac{x+y}{\sqrt2}}$$
what is the value of $k$?

Hi Olinguito!

Here is my attempt.
Take the total derivative:
$$dy\:=\:dx+\sqrt2\,e^{\frac{x+y}{\sqrt2}}\cdot \frac{dx+dy}{\sqrt2}$$
Substitute $dx=0$ to find where we have a vertical tangent:
$$ dy=\sqrt2\,e^{\frac{x+y}{\sqrt2}}\cdot \frac{dy}{\sqrt2}=dy\cdot e^{\frac{x+y}{\sqrt2}} \quad\Rightarrow\quad
y=-x$$
Substitute in the original equation:
$$-x\:=\:x+\sqrt2\,e^{\frac{x+(-x)}{\sqrt2}} \quad\Rightarrow\quad x=-\frac 12\sqrt 2$$
Thus:
$$k=-\frac 12\sqrt 2$$
 
  • #3
Great work! (Clapping) I have a different solution that does not involve calculus – again I’ll wait and see if anyone else wants to have a go.
 
  • #4
My solution.

A rotation of 45° clockwise about the origin sends the point $(x,y)$ to the point $(X,Y)$, where
$$\begin{pmatrix}X \\ Y\end{pmatrix}\ =\ \begin{pmatrix}\frac1{\sqrt2} & \frac1{\sqrt2} \\ -\frac1{\sqrt2} & \frac1{\sqrt2}\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}\ =\ \begin{pmatrix}\frac{x+y}{\sqrt2} \\ \frac{-x+y}{\sqrt2}\end{pmatrix}.$$
Thus the rotation takes the curve
$$\frac{-x+y}{\sqrt2}\ =\ e^{\frac{x+y}{\sqrt2}}$$
to the curve
$$Y\ =\ e^X.$$
The 45° tangent to the latter curve is the line $Y=X+1$. Rotating this 45° counterclockwise about the origin gives
$$\frac{-x+y}{\sqrt2}\ =\ \frac{x+y}{\sqrt2}\,+\,1$$
$\implies\ x\ =\ -\dfrac1{\sqrt2}$.
 
  • #5
Olinguito said:
If the line $x=k$ is tangent to the curve
$$\large y\:=\:x+\sqrt2\,e^{\frac{x+y}{\sqrt2}}$$
what is the value of $k$?

Differentiate both sides w.r.t. y and we get:$$\frac{dy}{dy}= \frac{dx}{dy}+\sqrt2\,e^{\frac{x+y}{\sqrt2}}\frac{1}{\sqrt 2}(\frac{dx}{dy}+\frac{dy}{dy})$$

And now since x=k is per/lar to the x-axis and tangent to the curve we must have : dx/dy =0 and the above formula becomes:
$$1=\,e^{\frac{x+y}{\sqrt2}}\Longrightarrow \frac{x+y}{\sqrt 2}=0\Longrightarrow x=-y$$ e.t.c,e.t.c
 

FAQ: Challenge problem Find k if x=k is tangent to the curve y=x+√(2).e^[(x+y)/√(2)]

What is the curve represented by the equation y=x+√(2).e^[(x+y)/√(2)]?

The curve represented by this equation is an exponential curve with a base of e raised to the power of x+y divided by the square root of 2, added to the value of x plus the square root of 2.

What does it mean for x=k to be tangent to the curve?

When x=k is tangent to a curve, it means that the line passing through the point (k, f(k)) is perpendicular to the tangent line of the curve at that point. In other words, the slope of the line passing through (k, f(k)) is equal to the slope of the tangent line at that point.

How can I find the value of k for which x=k is tangent to the curve?

To find the value of k, we can use the derivative of the curve and set it equal to the slope of the line passing through (k, f(k)). This will give us a system of equations which we can solve to find the value of k.

What is the significance of finding the value of k for which x=k is tangent to the curve?

Finding the value of k allows us to determine the point at which the curve has a unique tangent line. This can be useful in solving optimization problems or finding the minimum or maximum values of a function.

Can there be more than one value of k for which x=k is tangent to the curve?

Yes, there can be more than one value of k for which x=k is tangent to the curve. This occurs when the curve has a horizontal tangent line at multiple points, meaning the slope of the curve is equal to 0 at those points.

Similar threads

Replies
1
Views
916
Replies
4
Views
2K
Replies
12
Views
2K
Replies
2
Views
2K
Replies
1
Views
10K
Replies
18
Views
3K
Back
Top