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Question Video: Finding the Point on the Curve of a Function Involving Roots Such That the Tangent at This Point Is Perpendicular to a Given Line Mathematics

Determine the point at which the tangent to the curve √(π‘₯) + √(𝑦) = 15 is perpendicular to the line βˆ’2π‘₯ + 4𝑦 = 25.

06:26

Video Transcript

Determine the point at which the tangent to the curve root π‘₯ plus root 𝑦 is equal to 15 is perpendicular to the line negative two π‘₯ plus four 𝑦 is equal to 25.

In this question, we’re given a curve defined implicitly, root π‘₯ plus root 𝑦 is equal to 15, and the line negative two π‘₯ plus four 𝑦 is equal to 25. And we need to determine the point at which the tangent to this curve is perpendicular to the line. We can start by recalling the slope of the tangent line to the curve is given by its first derivative, d𝑦 by dπ‘₯, evaluated at a point. And for two lines to be perpendicular, either one is horizontal and the other is vertical or their slopes multiply to give negative one: π‘š sub one times π‘š sub two will be equal to negative one. And we can note we must be in the case where the product of the slopes is negative one. This is because we can directly find the slope of the straight line given in the question.

And one way of doing this is to rewrite our line in slope–intercept form. We add two π‘₯ to both sides of the equation and divide through by four. We see that 𝑦 is equal to π‘₯ over two plus 25 over four. Then, the slope of this straight line is the coefficient of π‘₯. We’ll call this π‘š sub two, which is equal to a half. We can then substitute π‘š sub two is equal to one-half into our equation involving perpendicular lines. Then, π‘š sub one will be the slope of the tangent line we want to find. This then gives us that π‘š sub one over two is equal to negative one. We can multiply through by two to see that π‘š sub one is negative two.

Therefore, we want to find the point on the curve where the slope of the tangent line at this point is negative two. Remember, this is given by d𝑦 by dπ‘₯ evaluated at the point. And since this curve is given implicitly, 𝑦 is not given as a function in π‘₯, we’re going to need to find an expression for d𝑦 by dπ‘₯ by using implicit differentiation.

Let’s start by rewriting the equation of this curve so that we can see the exponents of the variables. We have π‘₯ to the power of one-half plus 𝑦 to the power of one-half is equal to 15. And we’re going to want to differentiate both sides of this equation with respect to π‘₯ to find an expression for d𝑦 by dπ‘₯. We can differentiate this term by term. Let’s start with the first term. To differentiate π‘₯ to the power of one-half, we’re going to want to use the power rule for differentiation. We multiply by the exponent of π‘₯ and then reduce this exponent by one. The exponent of π‘₯ is one-half. So we multiply by one-half, and one-half minus one is negative one-half. This gives us one-half π‘₯ to the power of negative one-half.

And it’s worth noting we can simplify this. Multiplying by π‘₯ to the power of negative one-half is the same as dividing by π‘₯ to the power of one-half. And we can then rewrite π‘₯ to the power of one-half as root π‘₯. This gives us one over two root π‘₯. We now want to do the same with the second term on the left-hand side of this equation. We want to differentiate 𝑦 to the power of one-half with respect to π‘₯. We do need to be slightly careful however. Remember, 𝑦 itself is a function in π‘₯. So we need to differentiate this by using implicit differentiation. We can do this by using the chain rule. We start by differentiating 𝑦 to the power of one-half with respect to 𝑦.

We can do this by using the power rule for differentiation. However, we’ve already differentiated π‘₯ to the power of one-half with respect to π‘₯. Differentiating 𝑦 to the power of one-half with respect to 𝑦 will be the same. However, our variable will be 𝑦. This gives us one over two root 𝑦. Finally, to apply the chain rule, we also need to multiply this by d𝑦 by dπ‘₯. Finally, we need to differentiate the right-hand side of this equation. We note 15 is a constant, so its rate of change with respect to π‘₯ is zero. This gives us one over two root π‘₯ plus d𝑦 by dπ‘₯ times one over two root 𝑦 is equal to zero. This gives us an equation involving π‘₯, 𝑦, and d𝑦 by dπ‘₯. And remember, we’re looking for the point where the slope of the tangent line is negative two.

And since d𝑦 by dπ‘₯ gives us the slope of this tangent line, we can substitute d𝑦 by dπ‘₯ is equal to negative two. This then gives us that one over two root π‘₯ minus two over two root 𝑦 is equal to zero. We can then simplify this slightly. We can cancel the shared factor of two in the numerator and denominator of the second term. Therefore, any point on the curve which satisfies the equation one over two root π‘₯ minus one over root 𝑦 is equal to zero will have the slope of its tangent line equal to negative two. And in turn, this means it will be perpendicular to the line negative two π‘₯ plus four 𝑦 is equal to 25.

And to find the point on the curve which satisfies this equation, we’re going to need to rearrange the given equation for our curve. By subtracting root π‘₯ from both sides of the equation, we see that root 𝑦 is equal to 15 minus root π‘₯. If we then substitute this expression for root 𝑦 into our equation, we’ll find an equation entirely in terms of the variable π‘₯. This gives us that one over two root π‘₯ minus one over 15 minus root π‘₯ is equal to zero. Any value of π‘₯ which solves this equation will give us the potential points on the curve where the slope of the tangent line is perpendicular to the given line. And there’s a few different ways we could solve this equation. For example, we could add one over 15 minus root π‘₯ to both sides of the equation and then take the reciprocal of both sides of the equation.

However, we can also notice we’re subtracting two fractions and getting zero. The numerators of both fractions are the same. This means that the denominators of both fractions must also be equal. Using either method, we must have that two root π‘₯ is equal to 15 minus root π‘₯. And now we can just solve this equation for π‘₯. First, we add root π‘₯ to both sides of the equation. This gives us that three root π‘₯ is equal to 15. Then, we divide the equation through by three. We get root π‘₯ is equal to five. Finally, we square both sides of the equation. We get π‘₯ is equal to 25.

We now want to determine the 𝑦-coordinate of this point on the curve. We do this by substituting π‘₯ is equal to 25 into our rearrangement of the equation of the curve. Substituting π‘₯ is equal to 25 into this equation, we get root 𝑦 is equal to 15 minus root 25. And now we can solve this equation for 𝑦. First, we evaluate the right-hand side of the equation. We get that root 𝑦 is equal to 10. Then we square both sides of the equation. We get 𝑦 is equal to 100. Therefore, the π‘₯-coordinate of this point is 25 and its 𝑦-coordinate is 100. This point satisfies the equation of the curve, so it lies on our curve. And we’ve shown the slope of its tangent line at this point is negative two, which means it’s perpendicular to the given line.

And it’s worth noting we can double-check this. We can substitute π‘₯ is equal to 25 and 𝑦 is equal to 100 into our equation involving d𝑦 by dπ‘₯. If we did this, we could rearrange and solve for d𝑦 by dπ‘₯. And we would see it’s equal to negative two. In any case, we were able to show the point at which the tangent to the curve root π‘₯ plus root 𝑦 is equal to 15 is perpendicular to the line negative two π‘₯ plus four 𝑦 is equal to 25 is the point 25, 100.

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