The distance between the origin and the tangent to the curve $y = e^{2x} + x^2$ drawn at the point $x=0$ is?

Question

and the tangent to the curve $y = e^{2x} + x^2$ drawn at the point $x=0$ is ? I am not able to figure out what distance is he asking ? It's the distance between origin and the point where tangent cuts x or y axis Or something else ?

Answer 1

First find the tangent to the curve at $x=0$.

$y'(x) = 2e^{2x} + 2x$

$y'(0) = 2$

Next find the $y$-coordinate of the point on the curve for $x=0$.

$y(0) = 1$

Now find the equation of the tangent line.

$y-1 = 2(x-0)$

$y = 2x + 1$

Find the $y$ and $x$ intercepts, namely $y_0$ and $x_0$ of the tangent line.

$y_0 = y(0) = 1$.

$x_0$ occurs when $y = 0$ i.e. $x_0 = -\frac 12$.

Note that the points $(0,0), (x_0,0), (0,y_0)$ form a right triangle.

Find the area of this triangle one way, treating $|x_0|$ as the base and $|y_0|$ as the height.

Area $A = \frac 12 |x_0||y_0| = \frac 14$

Find the area of this triangle another way. Let the height $h$ be the required distance between the origin and the line, which means the base $b$ is the distance between $(x_0,0)$ and $(0,y_0)$.

$b = \sqrt{x_0^2 +y_0^2} = \frac 12{\sqrt 5}$

so $A = \frac 12 bh = \frac 14\sqrt 5 h$

Equating the two expressions for $A$ gives:

$h = \frac 1{\sqrt 5} = \frac 15 \sqrt 5$

Answer 2

Hint:

$$y_1=e^{2^x}2^x\ln2+2x$$

$$x=0,y=e,(y_1)_{x=0}=e\ln2$$

What is the equation of the tangent?

Do you know how to find the perpendicular distance of a straight line from a given point?

Answer 3

If the function is

$$y(x) = e^{2^x} + x^2$$ and not $$y(x) = e^{2x} + x^2,$$ then $$y(0) = e^{2^0} + 0 = e^1 = e$$

and we can build the first derivative (the slope of the tangent at some $x$)

$$y'(x) = \ln(2) ~ 2^x ~ e^{2^x} + 2x$$

and evaluate it at $x=0$ (the slope of the tangent at $x=0$)

$$y'(0) =\ln(2) ~ 2^0 ~ e^{2^0} + 0 =e~\ln(2).$$

The tangent line is (we have the slope and we have a point $(0,y(0))=(0,e)$ on the line):

$$t(x) = y'(0)~x+y(0)= e~\ln(2) ~ x +e$$

The distance between the origin $(0,0)$ and this tangent line can for example be calculated by building another line $p(x)$ which is perpendicular to the tangent line and passes through the origin. The equation for that line is

$$p(x) = -x/(e~\ln(2)) $$

The intersection of these two lines (can be obtained by solving $t(x)=p(x)$ and) is at $$(x,y)=(-(e~\ln(2))^2/((e~\ln(2))^2+1), (e~\ln(2))/((e~\ln(2))^2+1))$$

The distance between the origin and this intersection point is (pythagorean theorem, euclidian distance)

$$(e~\ln(2))/\sqrt{(e~\ln(2))^2+1}$$

Answer 4

In this question i have used the formula for distance between a point and the line .Am I right?(https://i.stack.imgur.com/lzGik.jpg)

Answer 5

Let $f(x)=e^{2x}+x^2$ $\in R$

$f'(x)=2e^{2x}+2x$, $f'(0)=2$, $f(0)=1$

Tangent line equation is: $y-f(0)=f'(0){x-0} \iff y-1=2x \iff 2x-y+1=0$

So, $(\epsilon): 2x-y+1$

The distance of origin from the line is given by: $d(O,\epsilon)=(\vert 2*0-1*0+1 \vert)/(\sqrt{2^2+(-1)^2})=1/\sqrt{5}=\sqrt{5}/5$

Therefore the distance is: $\sqrt{5}/5$