Question Video: Writing and Evaluating Exponential Functions to Model Exponential Decay in a Real-World Context Mathematics • Second Year of Secondary School

Video Transcript

A car’s value depreciates by 𝑟 percent every year. A new car costs 𝑃 dollars. Write a function that can be used to calculate 𝑉, the car’s value in dollars, after 𝑡 years. What is the value of 𝑟 for which the car’s value will be halved in three years? Give your answer to the nearest whole number.

For the first part of this question, we’re asked to find a function 𝑉, which is the value of a car, after 𝑡 years. We’re told that the car’s value depreciates by 𝑟 percent every year and that a new car costs 𝑃 dollars. We can model depreciation of this type with an exponential decay function, recalling that an exponential decay function can take the form 𝑉 of 𝑡 is equal to 𝑎 multiplied by 𝑏 raised to the power 𝑡, where 𝑏 is between zero and one. 𝑎 is the initial value. That’s 𝑉 at time 𝑡 is equal to zero. And 𝑏 is equal to uppercase 𝑅 plus one. That’s where 𝑅 is less than zero and represents the constant rate of change of 𝑉. And so, with this substitution, we have 𝑉 of 𝑡 equal to 𝑎 multiplied by one plus 𝑅 raised to the power 𝑡.

Now, since the car’s value 𝑉 depreciates by lowercase 𝑟 percent every year, we can represent the decay rate — that’s uppercase 𝑅 — as a percentage in terms of lowercase 𝑟, that is, negative lowercase 𝑟 over 100. We’re told that a new car costs 𝑃 dollars. So the initial value 𝑎 in our exponential decay function is equal to 𝑃. And so our function for the value of a car after 𝑡 years is 𝑉 of 𝑡 is equal to 𝑃 multiplied by one minus lowercase 𝑟 over 100 all raised to the power 𝑡.

Now, for the second part of the question, we want to find the value of 𝑟 for which the car’s value will be halved in three years. That is the yearly percentage decrease for which the car’s value will be halved in three years. Now, if the initial value of the car is 𝑃, then half of this is 𝑃 over two. And if the car’s value is halved in three years, then we have 𝑉 of three is equal to 𝑃 over two. But now if we substitute 𝑡 equal to three into our function 𝑉 of 𝑡, we have that 𝑉 of three is equal to 𝑃 multiplied by one minus 𝑟 over 100 raised to the power three.

And now equating our two expressions, we have an equation we can solve for 𝑟. First, dividing through by 𝑃, we’re left with one-half on the right-hand side and one minus 𝑟 over 100 to the power three on the left-hand side. Next, taking the cubed root on both sides, we have one minus 𝑟 over 100 is the cubed root of one-half. Now, multiplying both sides by negative one and adding one, we have 𝑟 over 100 is equal to one minus the cubed root of one-half. And multiplying both sides by 100 on our left-hand side, we isolate 𝑟. And so we have 𝑟 is equal to 100 multiplied by one minus the cubed root of one-half. And keying the right-hand side into our calculator gives us 𝑟 is equal 20.6299 and so on. To the nearest whole number then, we have 𝑟 is equal to 21 percent.

If a car’s value depreciates by 𝑟 percent every year and a new car costs 𝑃 dollars, we can calculate 𝑉, the car’s value, using an exponential decay function 𝑉 of 𝑡 is equal to 𝑃 multiplied by one minus 𝑟 over 100 raised to the power 𝑡. And the value of 𝑟 for which the car’s value will be halved in three years is 21 percent.