4. Given ln(x/y) + y^3 - 2x = -1.
A. Find the equation of the normal line to the curve ln(x/y) + y^3 - 2x = -1 at the point (1,1). (I got -2x+3)
B. Find the equation of a tangent line to the curve y=e^(x^2) that "also" passes through the point (1,0). You can approximate your final answer with your calculator after you've shown your work. (idk how to do this)
Let's rewrite things a bit to get
lnx - lny + y^3 - 2x = -1
Now,
A.
1/x - 1/y y' + 3y^2 y' - 2 = 0
y' (3y^2 - 1/y) = 2 - 1/x
y' = (2 - 1/x) / (3y^2 - 1/y)
so at (1,1), y' = (2-1)/(3-1) = 1/2
The normal there has slope -2, so
y-1 = -2(x-1)
y = -2x + 3
You are correct
B.
At any point on the curve, say, (m,e^(m^2)) the slope is 2m e^(m^2)
So now we have a point and a slope, so the equation is
y-0 = 2m e^(m^2) (x-1)
Now, we know that at x=m,
2m e^(m^2) (m-1) = e^m^2
2m(m-1) = 1
2m^2-2m-1 = 0
m = (1±√3)/2
So there are two lines:
y = (1+√3)e^(1+√3/2) (x-1)
y = (1-√3)e^(1-√3/2) (x-1)
You can see the graph of the first line and its tangency at
https://www.wolframalpha.com/input/?i=plot+y%3De%5Ex%5E2%2C+y+%3D+%281%2B%E2%88%9A3%29e%5E%281%2B%E2%88%9A3%2F2%29+%28x-1%29+for+0%3Cx%3C2
The other line is pretty hard to see, and just clutters things up if you plot it too.
A. Well, finding the equation of the normal line is no joke! But fear not, Clown Bot is here to make it fun. Let's start by finding the derivative of the given equation.
Taking the derivative of ln(x/y) with respect to x, we get 1/y.
Taking the derivative of y^3 with respect to x, we get 3y^2.
Taking the derivative of -2x with respect to x, we get -2.
Now, let's substitute the point (1,1) into our derivatives. Plugging in, we have:
1/y + 3(1)^2 - 2(1) = -1
Simplifying this equation, we find:
1/y + 3 - 2 = -1
1/y + 1 = -1
1/y = -2
y = -1/2
Great! Now we have the y-coordinate of the point where the normal line intersects the curve.
Next, let's find the corresponding x-coordinate at this point. We'll substitute y = -1/2 into the given equation:
ln(x/(-1/2)) + (-1/2)^3 - 2x = -1
Simplifying, we get:
ln(-2x) + (-1/8) - 2x = -1
Now, we just need to solve this equation for x. Once we have that value, we can plug it back into the original equation to find the corresponding y-value. The equation of the normal line will be in the form y = mx + b, where m is the slope and b is the y-intercept.
But Clown Bot's humor will take a break now, as solving this equation can be a bit tedious. So, grab your calculator and find an approximate solution for x. Then substitute it back into the original equation to find y, and you'll have all the information you need to form the equation of the normal line.
B. Now, let's move on to finding the equation of the tangent line to the curve y = e^(x^2) that passes through the point (1,0).
The equation of a tangent line can be expressed as y = mx + b, where m is the slope of the tangent line. In order to find m, we need to take the derivative of the curve equation and evaluate it at x = 1.
Taking the derivative of y = e^(x^2) with respect to x, we get:
dy/dx = 2x * e^(x^2)
Now, substitute x = 1 into the derivative equation:
dy/dx = 2(1) * e^(1^2)
dy/dx = 2 * e
So, the slope of the tangent line at x = 1 is 2e.
Now that we have the slope, we can use the point-slope form of a line to find the equation of the tangent line. Plugging in the coordinates of the point (1,0) and the slope 2e, we get:
y - 0 = 2e(x - 1)
Simplifying, we find:
y = 2ex - 2e
Voila! You now have the equation of the tangent line that also passes through the point (1,0). Don't forget to use your calculator to find an approximate value for e, and you're all set!
A. To find the equation of the normal line to the curve ln(x/y) + y^3 - 2x = -1 at the point (1,1), we need to find the slope of the normal line and the point at which it passes through.
1. First, let's differentiate the equation with respect to x to find the slope of the tangent line:
d/dx [ln(x/y) + y^3 - 2x] = d/dx [-1]
(1/y) * (dy/dx) + 3y^2 * (dy/dx) - 2 = 0
2. Now, substitute the coordinates of the point (1,1) into the derivative equation:
(1/1) * (dy/dx) + 3(1)^2 * (dy/dx) - 2 = 0
(dy/dx) + 3(dy/dx) - 2 = 0
4(dy/dx) - 2 = 0
4(dy/dx) = 2
(dy/dx) = 2/4
(dy/dx) = 1/2
3. The slope of the normal line is the negative reciprocal of the slope of the tangent line:
m_normal = -1/(dy/dx) = -1/(1/2) = -2
4. Now, we have the slope (-2) and the point (1,1). To find the equation of the line, use the point-slope form:
y - y1 = m(x - x1)
y - 1 = -2(x - 1)
y - 1 = -2x + 2
y = -2x + 3
Therefore, the equation of the normal line to the curve ln(x/y) + y^3 - 2x = -1 at the point (1,1) is y = -2x + 3.
B. To find the equation of a tangent line to the curve y=e^(x^2) that passes through the point (1,0), we need to find the derivative of the curve and then substitute the values into the point-slope form.
1. Find the derivative of y=e^(x^2):
dy/dx = d/dx [e^(x^2)]
dy/dx = 2x * e^(x^2)
2. Substituting the coordinate (1,0) into the derivative equation:
dy/dx = 2(1) * e^(1^2)
dy/dx = 2 * e^1
dy/dx = 2e
3. Now we can use the point-slope form to find the equation of the tangent line:
y - y1 = m(x - x1)
y - 0 = 2e(x - 1)
y = 2ex - 2e
Therefore, the equation of the tangent line to the curve y=e^(x^2) that passes through the point (1,0) is y = 2ex - 2e.
A. To find the equation of the normal line to the curve at the point (1, 1), we need to find the derivative of the curve and use it to find the slope of the tangent line at (1, 1). The normal line will have a slope that is the negative reciprocal of the slope of the tangent line.
Given the equation ln(x/y) + y^3 - 2x = -1, we can rewrite it as ln(x/y) = 2x - y^3 - 1.
Now, let's find the derivative of ln(x/y) with respect to x. Using the quotient rule, we have:
d/dx(ln(x/y)) = (1/y) * (1/y) * dy/dx * (x/y) - (1/y) * dx/dx * ln(x/y)
Simplifying this expression, we get:
d/dx(ln(x/y)) = (1/y^2) * (dy/dx * x - y)
Next, we need to find dy/dx. To do this, we differentiate both sides of the given equation with respect to x:
d/dx[ln(x/y) + y^3 - 2x] = d/dx(-1)
Using the sum rule and chain rule, we get:
(1/y^2) * (dy/dx * x - y) + 3y^2 * dy/dx - 2 = 0
Now, let's substitute the coordinates of the given point (1, 1) into this equation:
(1/y^2) * (dy/dx * 1 - 1) + 3(1)^2 * dy/dx - 2 = 0
Simplifying this expression, we have:
(1/y^2) * (dy/dx - 1) + 3dy/dx - 2 = 0
Let's solve this equation for dy/dx by isolating it:
(1/y^2 + 3)(dy/dx) = (1/y^2) - 1 + 2
Grouping the terms, we get:
(1/y^2 + 3)(dy/dx) = (1 - y^2 + 2y^2) / y^2
Simplifying further, we have:
(1/y^2 + 3)(dy/dx) = (y^2 + 1) / y^2
Now, let's plug in the coordinates of the point (1, 1):
(1/1^2 + 3)(dy/dx) = (1^2 + 1) / 1^2
(4)(dy/dx) = 2
dy/dx = 1/2
Therefore, the slope of the tangent line at the point (1, 1) is 1/2.
Since the normal line has the negative reciprocal slope of the tangent line, the slope of the normal line is -2.
Now we have the slope of the normal line (-2) and the point of tangency (1, 1). We can use the point-slope form of a line to find the equation of the normal line:
y - 1 = (-2)(x - 1)
Expanding and simplifying this equation, we get:
y - 1 = -2x + 2
y = -2x + 3
Therefore, the equation of the normal line to the curve at the point (1, 1) is y = -2x + 3.
B. To find the equation of a tangent line to the curve y = e^(x^2) that also passes through the point (1, 0), we need to find the derivative of the curve and use it to find the slope of the tangent line at (1, 0).
Given the equation y = e^(x^2), we can find dy/dx by differentiating both sides with respect to x:
d/dx(y) = d/dx(e^(x^2))
(dy/dx) = (d/dx(e^(x^2)))
Using the chain rule, we get:
(dy/dx) = e^(x^2) * 2x
Now, let's find the slope of the tangent line at the point (1, 0) by substituting x = 1 into the derivative:
(dy/dx) = e^(1^2) * 2(1)
(dy/dx) = 2e
Therefore, the slope of the tangent line at (1, 0) is 2e.
Using this slope and the point (1, 0), we can write the equation of the tangent line in point-slope form:
y - 0 = 2e(x - 1)
Simplifying this equation, we get:
y = 2ex - 2e
Therefore, the equation of the tangent line to the curve y = e^(x^2) that also passes through the point (1, 0) is y = 2ex - 2e.