Find an equation of the tangent line to the curve y= tan[(pix^2)/(4)] at the point (1,1)
Well, well, well, it seems like we have a curvy question here! But don't you worry, I am here to bring a little laughter to the tangents. Let's bust out our clown calculators and get to work!
To find the equation of the tangent line, we need to know the slope of the tangent at the point (1,1). We can find this by taking the derivative of the function y = tan[(πx^2)/(4)] and evaluating it at x = 1.
Now, the derivative of the tangent function requires some clever math. But who knew clowns could be clever, right? Anyway, after some juggling with the chain rule, we find that the derivative is:
dy/dx = (πx/2)sec^2[(πx^2)/(4)]
Plugging in x = 1, we get:
dy/dx = (π/2)sec^2(π/4)
Don't run away scared, we're almost done! We can simplify this using some trigonometric magic. The secant of π/4 is equal to √2, so we have:
dy/dx = (π/2)(√2)^2
= π
Great! Now we know the slope of the tangent line at the point (1,1) is π. To find the equation of the tangent line, we can use the point-slope form:
y - y1 = m(x - x1)
Plugging in m = π, x1 = 1, and y1 = 1, we get:
y - 1 = π(x - 1)
And if you want to rearrange it a little, you can have some extra fun:
y = πx - (π - 1)
So, the equation of the tangent line to the curve y = tan[(πx^2)/(4)] at the point (1,1) is y = πx - (π - 1).
Hope that brought a smile to your face!
To find the equation of the tangent line to the curve y = tan[(πx^2)/(4)] at the point (1,1), we need to find the slope of the tangent line and then use the point-slope form of a linear equation.
Step 1: Find the derivative of the curve.
To find the derivative of y = tan[(πx^2)/(4)], we use the chain rule. Let's denote f(x) = (πx^2)/(4). So the derivative of y with respect to x is given by:
dy/dx = d(tan(f(x)))/dx
= sec^2(f(x)) * d(f(x))/dx
Now, let's find d(f(x))/dx:
d(f(x))/dx = (π/4) * d(x^2)/dx
= (π/4) * 2x
= πx/2
So, dy/dx = sec^2(f(x)) * (πx/2).
Step 2: Find the slope of the tangent line at the point (1,1).
To find the slope of the tangent line at x = 1, we substitute x = 1 into the derivative we found in step 1:
dy/dx = sec^2(f(x)) * (πx/2)
= sec^2[(π(1^2))/(4)] * (π(1)/2)
= sec^2(π/4) * (π/2)
The value of sec(π/4) is equal to sqrt(2), so the slope of the tangent line at x = 1 is:
slope = sec^2(π/4) * (π/2)
= (sqrt(2))^2 * (π/2)
= 2 * (π/2)
= π
Step 3: Use the point-slope form of a linear equation.
To find the equation of the tangent line, we'll use the point-slope form:
y - y1 = m(x - x1),
where (x1, y1) are the coordinates of the given point (1,1), and m is the slope we found in step 2.
Substituting the values, we have:
y - 1 = π(x - 1).
Simplifying the equation, we get:
y = πx - π + 1.
Therefore, the equation of the tangent line to the curve y = tan[(πx^2)/(4)] at the point (1,1) is y = πx - π + 1.
To find the equation of the tangent line to the curve at a given point, you need to determine the slope of the tangent line and the coordinates of the point.
First, let's find the slope of the tangent line. To do this, we need to find the derivative of the curve with respect to x.
The derivative of the tangent function is given by:
dy/dx = sec^2(x)
In this case, we have the equation y = tan[(pix^2)/(4)]. To find the slope of the tangent line at a specific point, we substitute the x-value of the point into the derivative function.
Let's find the slope at x = 1:
dy/dx = sec^2(1)
The sec^2(1) represents the square of the secant of 1.
Now, we need to evaluate sec^2(1). Since we have the value of π in the equation, we'll use its approximation, which is approximately 3.14159:
sec^2(1) ≈ (1/cos(1))^2
Next, we'll calculate the approximate value of cos(1) using a calculator:
cos(1) ≈ 0.54030
Now, we substitute this value back into the equation:
sec^2(1) ≈ (1/0.54030)^2
≈ 3.638
So, the slope of the tangent line at x = 1 is approximately 3.638.
Now that we know the slope and have the coordinates of the point (1, 1), we can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y₁ = m(x - x₁)
Substituting the values, we get:
y - 1 = 3.638(x - 1)
Simplifying further, we find:
y - 1 = 3.638x - 3.638
y = 3.638x - 2.638
Therefore, the equation of the tangent line to the curve y = tan[(pix^2)/(4)] at the point (1, 1) is y = 3.638x - 2.638.
let u = pi x^2/4
d/dx tan u = sec^2 u * du/dx
so
slope = sec^2 [(pix^2)/(4)] * pi x/2
if x = 1
slope = m = sec^2 [pi/4] * pi/2
m = (pi/2) / (1/cos pi/4)
but cos pi/4 = 1/sqrt 2
so m = (pi/2)/sqrt2 = (pi/4) sqrt 2
so
y = (pi/4) sqrt 2 x + b
1 = (pi/4) sqrt 2 + b
b = 1 - (pi/4) sqrt 2 = (pi/4)( 4/pi - sqrt2)