I need some help with my maths questions please!
What angle does the tangent to the curve y=x^4-x^3+7x make with the positive direction of the x-axis at the point where x=2?
Find any x values for which the gradient of the tangent to the curve y=x^2-3x-1 is parallel to the line x-2y-1=0.
recall that y' is the slope of the tangent line. The slope is the tanθ, so
y'(x) = 4x^3-3x^2+7
y'(2) = 4*8-3*4+7 = 27 = tanθ
θ = 87.9°
The line x-2y-1=0 has slope 1/2
So, where is y' 1/2?
y' = 2x-3
2x-3 = 1/2 when x = 7/4
Of course, I'll be glad to help you with your math questions!
Let's start with the first question about the angle of the tangent to the curve at a specific point.
To find the angle, we need to know the slope (gradient) of the tangent line at that point. The slope of the tangent can be found by taking the derivative of the curve's equation with respect to x.
So for the first curve, y = x^4 - x^3 + 7x, we need to find its derivative. Let's call the derivative dy/dx.
Now, differentiate each term of the equation and combine like terms:
dy/dx = 4x^3 - 3x^2 + 7
Next, substitute the x-value of the point of interest into the derivative equation. In this case, the point is x = 2.
dy/dx = 4(2^3) - 3(2^2) + 7
= 32 - 12 + 7
= 27
Now we have the slope of the tangent at that point, which is 27.
To find the angle the tangent makes with the positive x-axis, we can use the formula:
angle = arctan(m)
Where 'm' is the slope of the tangent line. In this case, m = 27.
angle = arctan(27)
≈ 86.41 degrees
So the angle between the tangent and the positive x-axis at the point (x=2) is approximately 86.41 degrees.
Moving on to the second question about finding x-values where the gradient of the tangent is parallel to a given line:
The equation of the given line is x - 2y - 1 = 0.
First, we need to find the slope (gradient) of this line. Rewrite the equation in slope-intercept form, y = mx + c, where 'm' represents the slope. Rearranging the equation, we have:
2y = x - 1
y = (1/2)x - 1/2
Comparing this equation with y = mx + c, we see that the slope of the line is 1/2.
To find the x-values where the gradient of the tangent is parallel to the given line, we need to find where the derivative of the curve y = x^2 - 3x - 1 is equal to the slope of the line (1/2).
So, differentiate the equation of the curve:
dy/dx = 2x - 3
Now, set the derivative equal to the slope of the line:
2x - 3 = 1/2
Solve for x:
2x = 1/2 + 3
2x = 7/2
x = 7/4
Therefore, the x-value for which the gradient of the tangent to the curve is parallel to the line x - 2y - 1 = 0 is x = 7/4.
I hope this explanation helps you understand how to solve these math problems. Let me know if you have any further questions!