The normal to the curve at #y=asqrtx+b/sqrtx# where a and b are constants is #4x+y=22# at the point where x=4 and how do you find a and b?
We know that the slope of normal to the curve
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To find the values of a and b, we need to use the given information about the normal to the curve.
First, let's find the derivative of the curve equation y = a√x + b/√x with respect to x.
dy/dx = (1/2) * a * x^(-1/2) - (1/2) * b * x^(-3/2)
Next, we know that the normal to the curve is perpendicular to the tangent at the given point (x=4).
The slope of the tangent at x=4 can be found by substituting x=4 into the derivative:
dy/dx = (1/2) * a * 4^(-1/2) - (1/2) * b * 4^(-3/2)
Now, we can determine the slope of the normal by taking the negative reciprocal of the tangent's slope:
m_normal = -1 / [(1/2) * a * 4^(-1/2) - (1/2) * b * 4^(-3/2)]
Since the normal passes through the point (4, y), we can substitute these values into the equation of the normal line (4x + y = 22) to find the value of y:
4 * 4 + y = 22
Solving this equation will give us the value of y.
Finally, we can substitute the values of x=4 and y into the derivative equation and solve for a and b.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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