Video: Finding the Values of x at Which Two Curves Have Perpendicular Tangent Lines

At which of the following values of π‘₯ do the graphs of 𝑦 = 2π‘₯Β² and 𝑦 = βˆ’2√(π‘₯) have perpendicular tangent lines? [A] βˆ’2 [B] 1 [C] 4 [D] None of the choices are valid.

06:36

Video Transcript

At which of the following values of π‘₯ do the graphs of 𝑦 equals two π‘₯ squared and 𝑦 equals negative two square root of π‘₯ have perpendicular tangent lines? Is it (a) π‘₯ equals negative two, (b) π‘₯ equals one, (c) π‘₯ equals four, or (d) none of the choices are valid?

In this question, we are given two curves, 𝑦 equals two π‘₯ squared and 𝑦 equals negative two square root of π‘₯. For clarity, let’s relabel them as 𝑦 one equals two π‘₯ squared and 𝑦 two equals negative two square root of π‘₯. Since we’re allowed to use a graphical calculator for this question, let’s see if we can draw the two curves for a better understanding of the question. The curve 𝑦 one equals two π‘₯ squared is a parabola with an upward opening. The curve 𝑦 two equals negative two square root of π‘₯ is half of a parabola with an opening towards the right side. We are required to determine at which of the given values of π‘₯ do the graphs of 𝑦 one and 𝑦 two have perpendicular tangent lines.

Let’s try to figure out what this means. First of all, note that the correct answer to the question cannot be option (a), which is π‘₯ equals negative two, as the square root function in the curve 𝑦 two and, hence, the function 𝑦 two itself are not defined when π‘₯ is a negative number. Suppose the correct answer to the question was option (b), π‘₯ equals one. Then, it would mean that the tangent lines at π‘₯ equals one on the graphs of 𝑦 one and 𝑦 two, as drawn on the diagram, are perpendicular, i.e., at an angle of 90 degrees to each other.

In order to find out whether or not the tangent lines at π‘₯ equals one to the curves 𝑦 one and 𝑦 two are actually perpendicular to each other, we can use a criteria that we know for perpendicular lines that are not horizontal or vertical. We have that if the lines 𝑙 one and 𝑙 two are perpendicular, then π‘š one multiplied by π‘š two equals negative one, where π‘š one and π‘š two are gradients of 𝑙 one and 𝑙 two, respectively. This also holds the other way around, i.e., if π‘š one multiplied by π‘š two equals negative one, where π‘š one and π‘š two are gradients of lines 𝑙 one and 𝑙 two, respectively. Then, the lines 𝑙 one and 𝑙 two are perpendicular.

Remember that the gradient of a line is a constant function. And the gradient of a tangent line, which is not vertical, is the constant function obtained by evaluating the gradient function of the curve to which the line is tangent at the tangent point. So, if we label the tangent lines to the curves 𝑦 one and 𝑦 two at π‘₯ equals one by 𝑙 one and 𝑙 two, respectively. Then π‘š one, the gradient of 𝑙 one, is equal to the gradient function of the curve 𝑦 one, which is just its first derivative with respect to π‘₯ evaluated at π‘₯ equals one. Similarly, π‘š two, the gradient of 𝑙 two, is equal to the gradient function of the curve 𝑦 two, which is the first derivative of 𝑦 two with respect to π‘₯ evaluated at π‘₯ equals one.

Using the power rule for differentiation, the first derivative of 𝑦 one with respect to π‘₯ is equal to the coefficient two multiplied by the exponent two multiplied by π‘₯ to the power of the exponent reduced by one, which is two minus one. This simplifies to four π‘₯. Similarly, writing 𝑦 two equals negative two square root of π‘₯ as negative two π‘₯ to the power of a half. We have that the first derivative of 𝑦 two with respect to π‘₯ is equal to negative two multiplied by a half multiplied by π‘₯ to the power of a half minus one. Which simplifies to negative π‘₯ to the power of negative a half. We can rewrite this as negative one over π‘₯ to the power of a half.

We can clearly see from the diagram that 𝑙 one and 𝑙 two are not horizontal or vertical tangent lines. And so, if 𝑙 one and 𝑙 two are perpendicular. Then by the criteria for perpendicular lines, you must have that π‘š one β€” which equals four π‘₯ evaluated at π‘₯ equals one, which is four. And π‘š two β€” which equals negative one over π‘₯ to the power of a half evaluated at π‘₯ equals one, which equals negative one β€” multiply to make negative one. We can clearly see that this is not the case, as four times negative one is negative four, which is not equal to negative one. So, we deduce that 𝑙 one and 𝑙 two are not perpendicular, which means that the graphs of the curves 𝑦 one and 𝑦 two do not have perpendicular tangent lines at π‘₯ equals one. This means that option (b) is not the correct answer to the question.

Now, we could repeat this procedure by replacing π‘₯ equals one with π‘₯ equals four in order to determine the correct answer to the question. However, let’s see if we can find the value of π‘₯ at which the graphs of 𝑦 one and 𝑦 two have perpendicular tangent lines. In order to do this, all we need to do is solve the equation four π‘₯ multiplied by negative one over π‘₯ to the power of a half equals negative one. As by the criteria for perpendicular lines. The statement that the tangent lines to the curves 𝑦 one and 𝑦 two at a particular π‘₯-value are perpendicular is equivalent to the statement that the gradients of the tangent lines. Which are obtained by evaluating the first derivatives of 𝑦 one and 𝑦 two at the tangent point, multiply to equal negative one.

Since the criteria of our perpendicular lines that we are using does not hold for perpendicular lines that are horizontal or vertical. We are assuming that the solutions π‘₯ to the equation do not correspond to horizontal or vertical tangent lines on 𝑦 one and 𝑦 two. We can see from the diagram that 𝑦 one and 𝑦 two do indeed have horizontal and vertical perpendicular tangent lines at π‘₯ equals zero. And so, we are assuming that π‘₯ does not equal to zero when solving this equation. Canceling the negative sign from both sides of the equation, we obtain four π‘₯ over π‘₯ to the power of a half equals one.

By the quotient of powers rule, this implies that four times π‘₯ to the power of a half equals one. As a result, we obtain that π‘₯ to the power of a half equals one over four. Squaring both sides of the equation, we obtain π‘₯ equals one over 16. So, the graphs of 𝑦 equals two π‘₯ squared and 𝑦 equals negative two square root of π‘₯ have perpendicular tangent lines at π‘₯ equals one over 16 and π‘₯ equals zero. So, the correct answer to the question is option (d). None of the choices (a), (b), or (c) are valid.

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