Video: Rectilinear Motion Using Trigonometric Functions

At what value(s) of π‘₯ do the graphs of 𝑦 = βˆ’π‘’^(βˆ’π‘₯) and 𝑦 = βˆ’(1/2)π‘₯Β² + 3π‘₯ βˆ’ 7 have parallel tangents?

03:17

Video Transcript

At what value or values of π‘₯ do the graphs of 𝑦 equals negative 𝑒 to the power of negative π‘₯ and 𝑦 equals negative one-half π‘₯ squared plus three π‘₯ minus seven have parallel tangents?

In order for the tangents to each curve to be parallel, we know that the slope of their respective tangents must be equal. We also know that, to find the slope of a tangent to a curve given by 𝑦 is equal to some function of π‘₯, we need to find its first derivative. That’s d𝑦 by dπ‘₯.

So let’s begin by differentiating our first equation with respect to π‘₯. We need to differentiate negative 𝑒 to the power of negative π‘₯. Well, we know that when we differentiate 𝑒 to the power of π‘Žπ‘₯, where π‘Ž is some real constant, we get π‘Ž times 𝑒 to the power of π‘Žπ‘₯. Well, in this case, π‘Ž is really equal to negative one. So the derivative of negative 𝑒 to the power of negative π‘₯ with respect to π‘₯ is negative one times 𝑒 to the power of negative π‘₯, which is simply 𝑒 to the power of negative π‘₯.

Let’s repeat this process for the equation 𝑦 equals negative a half π‘₯ squared plus three π‘₯ minus seven. This is a polynomial function with three terms. And we can actually differentiate term by term. We begin by differentiating negative a half π‘₯ squared. To differentiate negative one-half times π‘₯ to the power of two, we multiply the entire term by the exponent and then reduce the exponent by one. So we get two times negative one-half times π‘₯ to the power of one. But π‘₯ to the power of one is just π‘₯. Two times negative one-half is negative one.

So our first term differentiates to negative π‘₯. The derivative of three π‘₯ is simply three, and the derivative of negative seven is zero. So when we differentiate our second equation with respect to π‘₯, we get negative π‘₯ plus three. Remember, we said that, for the tangents to be parallel, their slopes must be equal. And so we can say that we need to find the values of π‘₯ such that 𝑒 to the power of negative π‘₯ is equal to negative π‘₯ plus three. And this isn’t a particularly nice equation to solve. But we are allowed to use our calculators here.

What we’re going to do is subtract 𝑒 to the power of negative π‘₯ from both sides of our equation. So zero is equal to negative π‘₯ plus three minus 𝑒 to the power of negative π‘₯. We’re then going to use our calculators to sketch the curve of 𝑦 equals negative π‘₯ plus three minus 𝑒 to the power of negative π‘₯. Then the solutions to the equation zero equals negative π‘₯ plus three minus 𝑒 to the power of negative π‘₯ will be the values of the π‘₯-intercepts. Remember, this is because the π‘₯-axis has the equation 𝑦 equals zero.

Now it’s actually very difficult for us to see the entire shape of this curve. But if we look closely, we see that it intersects the π‘₯-axis at negative 1.5052 and so on and at 2.9475 and so on. And so, correct to three decimal places, the two solutions to the equation 𝑒 to the power of negative π‘₯ equals negative π‘₯ plus three are π‘₯ equals 2.948 and π‘₯ equals negative 1.505.

Now we said that the values of π‘₯ that satisfy this equation are the values of π‘₯ that the graphs of 𝑦 equals negative 𝑒 to the power of negative π‘₯ and 𝑦 equals negative one half π‘₯ squared plus three π‘₯ minus seven have parallel tangents. So we’ve answered the question. They are 2.948 and negative 1.505.

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